Samuel Dominic Chukwuemeka (SamDom For Peace)

For in GOD we live, and move, and have our being. - Acts 17:28

The Joy of a Teacher is the Success of his Students. - Samuel Chukwuemeka



Solved Examples - Literal Equations

For ACT Students
The ACT is a timed exam...$60$ questions for $60$ minutes
This implies that you have to solve each question in one minute.
Some questions will typically take less than a minute a solve.
Some questions will typically take more than a minute to solve.
The goal is to maximize your time. You use the time saved on those questions you solved in less than a minute, to solve the questions that will take more than a minute.
So, you should try to solve each question correctly and timely.
So, it is not just solving a question correctly, but solving it correctly on time.
Please ensure you attempt all ACT questions.
There is no "negative" penalty for any wrong answer.

For JAMB and CMAT Students
Calculators are not allowed. So, the questions are solved in a way that does not require a calculator.

As applicable:
Solve each literal equation for the specified variable.
Each literal equation is a formula.
Name the formula and the subject/content area as applicable.
Simplify completely.
Show all work.

(1.) $V = lwh$ for $w$


$V = lwh$
Volume of a rectangle: Mathematics/Geometry

We need to isolate $w$

$V = l * w * h$
Swap. Let the LHS = RHS, and the RHS = LHS

$l * w * h = V$
Divide both sides by $l * h$

$ \dfrac{l * w * h}{l * h} = \dfrac{V}{l * h} \\[5ex] w = \dfrac{V}{lh} $
(2.) $F = \dfrac{mv}{t}$ for $t$


$F = \dfrac{mv}{t}$

Force of a body: Physics/Mechanics

We need to isolate $t$

$ F = \dfrac{mv}{t} \\[3ex] Remove\:\: the\:\: fraction.\:\: Find\:\: the\:\: LCD \\[3ex] LCD = t \\[3ex] $ Multiply both sides by $t$

$ t * F = t * \dfrac{mv}{t} \\[5ex] t * F = mv \\ $ Divide both sides by $F$

$ t = \dfrac{mv}{F} $
(3.) $P = a + b + c$ for $c$


$P = a + b + c$
Perimeter of a triangle: Mathematics/Geometry,Trigonometry

We need to isolate $c$

$P = a + b + c$
Swap. Let the LHS = RHS, and the RHS = LHS

$a + b + c = P$
Subtract $a$ and $b$ from both sides.

$ a + b + c - a - b = P - a - b \\[2ex] c = P - a - b $
(4.) $c^2 = a^2 + b^2$ for $a$


$c^2 = a^2 + b^2$
Pythagorean Theorem: Mathematics/Geometry,Trigonometry

We need to isolate $a$

$c^2 = a^2 + b^2$

Swap. Let the LHS = RHS, and the RHS = LHS

$a^2 + b^2 = c^2$

Subtract $b^2$ from both sides

$a^2 = c^2 - b^2$

Apply the Square Root property.

$ a = \pm \sqrt{c^2 - b^2} \\[3ex] a = \pm \sqrt{(c + b)(c - b)} $
(5.) $A = \dfrac{1}{2}h(l_1 + l_2)$ for $l_2$


$A = \dfrac{1}{2}h(l_1 + l_2)$

Area of a trapezoid/trapezium: Mathematics/Geometry

We need to isolate $l_2$

$A = \dfrac{1}{2}h(l_1 + l_2)$

Swap. Let the LHS = RHS, and the RHS = LHS

$
\dfrac{1}{2}h(l_1 + l_2) = A \\[5ex] Remove\:\: the\:\: fraction.\:\: Find\:\: the\:\: LCD \\[3ex] LCD = 2 \\[3ex] $ Multiply both sides by $2$

$ 2 * \dfrac{1}{2}h(l_1 + l_2) = 2 * A \\[5ex] h(l_1 + l_2) = 2 * A \\ $ Divide both sides by $h$

$l_1 + l_2 = \dfrac{2A}{h}$

Subtract $l_1$ from both sides

$ l_2 = \dfrac{2A}{h} - l_1 \\[5ex] l_2 = \dfrac{2A}{h} - \dfrac{hl_1}{h} \\[5ex] l_2 = \dfrac{2A - hl_1}{h} $
(6.) $A = \dfrac{1}{2}h(l_1 + l_2)$ for $h$


$A = \dfrac{1}{2}h(l_1 + l_2)$

Area of a trapezoid/trapezium: Mathematics/Geometry

We need to isolate $h$

$A = \dfrac{1}{2}h(l_1 + l_2)$

Swap. Let the LHS = RHS, and the RHS = LHS

$
\dfrac{1}{2}h(l_1 + l_2) = A \\[5ex] Remove\:\: the\:\: fraction.\:\: Find\:\: the\:\: LCD \\[3ex] LCD = 2 \\[3ex] $ Multiply both sides by $2$

$ 2 * \dfrac{1}{2}h(l_1 + l_2) = 2 * A \\[5ex] h(l_1 + l_2) = 2 * A \\ $ Divide both sides by $(l_1 + l_2)$

$ h = \dfrac{2A}{l_1 + l_2} $
(7.) $s = ut + \dfrac{1}{2}at^2$ for $u$


$s = ut + \dfrac{1}{2}at^2$

Newton's Laws of Motion: Physics/Mechanics

We need to isolate $u$

$s = ut + \dfrac{1}{2}at^2$

Swap. Let the LHS = RHS, and the RHS = LHS

$ ut + \dfrac{1}{2}at^2 = s \\[5ex] Remove\:\: the\:\: fraction.\:\: Find\:\: the\:\: LCD \\[3ex] LCD = 2 \\[3ex] $ Multiply both sides by $2$

$ 2 * \left(ut + \dfrac{1}{2}at^2\right) = 2 * s \\[5ex] 2 * ut + 2 * \dfrac{1}{2}at^2 = 2s \\[5ex] 2ut + at^2 = 2s $
Subtract $at^2$ from both sides

$2ut = 2s - at^2$

Divide both sides by $2t$

$ u = \dfrac{2s - at^2}{2t} $
(8.) $s = ut + \dfrac{1}{2}at^2$ for $t$


$s = ut + \dfrac{1}{2}at^2$

Newton's Laws of Motion: Physics/Mechanics

We need to isolate $t$

$s = ut + \dfrac{1}{2}at^2$

Swap. Let the LHS = RHS, and the RHS = LHS

$ ut + \dfrac{1}{2}at^2 = s \\[5ex] Remove\:\: the\:\: fraction.\:\: Find\:\: the\:\: LCD \\[3ex] LCD = 2 \\[3ex] $ Multiply both sides by $2$

$ 2 * \left(ut + \dfrac{1}{2}at^2\right) = 2 * s \\[5ex] 2 * ut + 2 * \dfrac{1}{2}at^2 = 2s \\[5ex] 2ut + at^2 = 2s $
This is a quadratic equation in $t$.
Put it in standard form.

$ at^2 + 2ut = 2s \\[3ex] at^2 + 2ut - 2s = 0 $
Apply the Quadratic Formula

$ ax^2 + bx + c = 0 \\[3ex] x = \dfrac{-b \pm \sqrt{b^2 - 4ac}}{2a} \\[5ex] Compare \\[2ex] a = a, b = 2u, c = -2s \\[2ex] t = \dfrac{-b \pm \sqrt{b^2 - 4ac}}{2a} \\[5ex] t = \dfrac{-2u \pm \sqrt{(2u)^2 - 4(a)(-2s)}}{2a} \\[5ex] t = \dfrac{-2u \pm \sqrt{4u^2 - (-8as)}}{2a} \\[5ex] t = \dfrac{-2u \pm \sqrt{4u^2 + 8as}}{2a} \\[5ex] t = \dfrac{-2u \pm \sqrt{4(u^2 + 2as)}}{2a} \\[5ex] t = \dfrac{-2u \pm \sqrt{4} * \sqrt{u^2 + 2as}}{2a} \\[5ex] t = \dfrac{-2u \pm 2 * \sqrt{u^2 + 2as}}{2a} \\[5ex] t = \dfrac{2\left(-u \pm \sqrt{u^2 + 2as}\right)}{2a} \\[5ex] t = \dfrac{-u \pm \sqrt{u^2 + 2as}}{a} \\[5ex] $
(9.) $A = P\left(1 + \dfrac{r}{m}\right)^{mt}$ for $t$


$A = P\left(1 + \dfrac{r}{m}\right)^{mt}$

Compound Interest: Mathematics/Algebra/Mathematics of Finance

We need to isolate $t$

$ A = P\left(1 + \dfrac{r}{m}\right)^{mt} \\[5ex] Swap \\[3ex] P\left(1 + \dfrac{r}{m}\right)^{mt} = A \\[5ex] Divide\:\:both\:\:sides\:\:by\:\:P \\[3ex] \left(1 + \dfrac{r}{m}\right)^{mt} = \dfrac{A}{P} \\[5ex] Take\:\:the\:\:logarithm\:\:of\:\:both\:\:sides \\[3ex] \log\left(1 + \dfrac{r}{m}\right)^{mt} = \log\left(\dfrac{A}{P}\right) \\[5ex] mt\log\left(1 + \dfrac{r}{m}\right) = \log\left(\dfrac{A}{P}\right) \\[5ex] Divide\:\:both\:\:sides\:\:by\:\:m\log\left(1 + \dfrac{r}{m}\right) \\[5ex] t = \dfrac{\log\left(\dfrac{A}{P}\right)}{m\log\left(1 + \dfrac{r}{m}\right)} $
(10.) $A = 2(lw + wh + lh)$ for $l$


$A = 2(lw + wh + lh)$
Surface Area of a Rectangular Prism: Mathematics/Geometry

We need to isolate $l$

$ A = 2(lw + wh + lh) A = 2lw + 2wh + 2lh \\[3ex] $ Swap. Let the LHS = RHS, and the RHS = LHS

$2lw + 2wh + 2lh = A \\[3ex]$ Collect the "like terms" in "l" to be only on the $LHS$

$2lw + 2lh = A - 2wh \\[3ex]$ Factor by GCF.

$l(2w + 2h) = A - 2wh \\[3ex]$ Divide both sides by $(2w + 2h)$

$ l = \dfrac{A - 2wh}{2w + 2h} \\[5ex] OR \\[3ex] l = \dfrac{A - 2wh}{2(w + h)} $
(11.) $V = \dfrac{4}{3}\pi r^3$ for $r$


$V = \dfrac{4}{3}\pi r^3$

Volume of a sphere: Mathematics/Geometry

We need to isolate $r$

$V = \dfrac{4}{3}\pi r^3 \\[5ex]$ Swap. Let the LHS = RHS, and the RHS = LHS

$\dfrac{4}{3}\pi r^3 = V \\[5ex]$ Remove the fraction. Find the $LCD$
$LCD = 3$
Multiply both sides by the $LCD$

$ 3 * \dfrac{4}{3}\pi r^3 = 3 * V \\[5ex] 4\pi r^3 = 3V \\[3ex] $ Divide both sides by $4\pi$

$ \dfrac{4\pi r^3}{4\pi} = \dfrac{3V}{4\pi} \\[5ex] r^3 = \dfrac{3V}{4\pi} \\[5ex] $ Multiply both expoents by $\dfrac{1}{3}$

In other words, take the cube root of both sides

$ \sqrt[3]{r^3} = \sqrt[3]{\dfrac{3V}{4\pi}} \\[5ex] r = \sqrt[3]{\dfrac{3V}{4\pi}} $
(12.) $APY = \left(1 + \dfrac{r}{m}\right)^m - 1$ for $r$


$APY = \left(1 + \dfrac{r}{m}\right)^m - 1$

Annual Percentage Yield (APY): Mathematics/Algebra/Mathematics of Finance

We need to isolate $r$

$APY = \left(1 + \dfrac{r}{m}\right)^m - 1 \\[5ex]$ Swap. Let the LHS = RHS, and the RHS = LHS

$\left(1 + \dfrac{r}{m}\right)^m - 1 = APY \\[5ex]$ Add $1$ to both sides.

$\left(1 + \dfrac{r}{m}\right)^m = APY + 1 \\[5ex]$ Multiply both exponents by $\dfrac{1}{m}$

$ \left(1 + \dfrac{r}{m}\right)^{m * \dfrac{1}{m}} = (APY + 1)^{\dfrac{1}{m}} \\[5ex] 1 + \dfrac{r}{m} = (APY + 1)^{\dfrac{1}{m}} \\[5ex] $ Subtract $1$ from both sides

$\dfrac{r}{m} = (APY + 1)^{\dfrac{1}{m}} - 1 \\[5ex]$ Remove the fraction. Find the $LCD$
$LCD = m$
Multiply both sides by the $LCD$

$ m * \dfrac{r}{m} = m * \left((APY + 1)^{\dfrac{1}{m}} - 1\right) \\[5ex] r = m\left[(APY + 1)^{\dfrac{1}{m}} - 1\right] \\[5ex] OR \\[3ex] r = m\left(\sqrt[m]{APY + 1} - 1\right) $
(13.) Celsius - Fahrenheit temperature conversion
$F = \dfrac{9}{5}C + 32$ for $C$


$F = \dfrac{9}{5}C + 32$ for $C$
Celsius - Fahrenheit temperature conversion and Inverse Functions: Mathematics/Algebra

We need to isolate $C$

$F = \dfrac{9}{5}C + 32 \\[5ex]$ Swap. Let the LHS = RHS, and the RHS = LHS

$\dfrac{9}{5}C + 32 = F \\[5ex]$ Remove the fraction. Find the $LCD$
$LCD = 5$
Multiply both sides by the $LCD$

$ 5\left(\dfrac{9}{5}C + 32\right) = 5 * F \\[5ex] 5 * \dfrac{9}{5}C + 5(32) = 5F \\[5ex] 9C + 160 = 5F \\[3ex] $ Subtract $160$ from both sides.

$9C = 5F - 160 \\[3ex]$ Divide both sides by $9$

$ \dfrac{9C}{9} = \dfrac{5F - 160}{9} \\[5ex] C = \dfrac{5(F - 32)}{9} $
(14.) Celsius - Fahrenheit temperature conversion
$C = \dfrac{5}{9}(F - 32)$ for $F$


$C = \dfrac{5}{9}(F - 32)$ for $F$
Celsius - Fahrenheit temperature conversion and Inverse Functions: Mathematics/Algebra

We need to isolate $F$

$C = \dfrac{5}{9}(F - 32) \\[5ex]$ Swap. Let the LHS = RHS, and the RHS = LHS

$\dfrac{5}{9}(F - 32) = C \\[5ex]$ Remove the fraction. Find the $LCD$
$LCD = 9$
Multiply both sides by the $LCD$

$ 9\left(\dfrac{5}{9}(F - 32)\right) = 9 * C \\[5ex] 9 * \dfrac{5}{9} * (F - 32) = 9C \\[5ex] 5(F - 32) = 9C \\[3ex] $ Divide both sides by $5$

$ \dfrac{5(F - 32)}{5} = \dfrac{9C}{5} \\[5ex] F - 32 = \dfrac{9C}{5} \\[5ex] $ Add $32$ to both sides

$ F = \dfrac{9C}{5} + 32 \\[5ex] F = \dfrac{9C}{5} + \dfrac{160}{5} \\[5ex] F = \dfrac{9C + 160}{5} $
(15.) $SGS_n = \dfrac{a(r^n - 1)}{r - 1}$ for $n$


$SGS_n = \dfrac{a(r^n - 1)}{r - 1}$ for $n$
Sum of the $n$ terms of a Geometric Sequence: Mathematics/Algebra

We need to isolate $n$

$SGS_n = \dfrac{a(r^n - 1)}{r - 1} \\[5ex]$ Swap. Let the LHS = RHS, and the RHS = LHS

$\dfrac{a(r^n - 1)}{r - 1} = SGS_n \\[5ex]$ Remove the fraction. Find the $LCD$
$LCD = r - 1$
Multiply both sides by the $LCD$

$ (r - 1) * \left[\dfrac{a(r^n - 1)}{r - 1}\right] = (r - 1) * SGS_n \\[5ex] a(r^n - 1) = SGS_n(r - 1) \\[3ex] Divide\:\: both\:\: sides\:\: by\:\: a \\[3ex] r^n - 1 = \dfrac{SGS_n(r - 1)}{a} \\[5ex] r^n = \dfrac{SGS_n(r - 1)}{a} + 1 \\[5ex] Take\:\: the\:\: logarithm\:\: of\:\: both\:\: sides \\[3ex] \log{r^n} = \log{\left[\dfrac{SGS_n(r - 1)}{a} + 1\right]} \\[5ex] n \log r = \log{\left[\dfrac{SGS_n(r - 1)}{a} + 1\right]} \\[5ex] Divide\:\: both\:\: sides\:\: by\:\: \log r \\[3ex] n = \dfrac{\log{\left[\dfrac{SGS_n(r - 1)}{a} + 1\right]}}{\log r} $
(16.) $SAS_n = \dfrac{n}{2}\left[2a + d(n - 1)\right]$ for $n$


$SAS_n = \dfrac{n}{2}\left[2a + d(n - 1)\right]$ for $n$
Sum of the $n$ terms of an Arithmetic Sequence: Mathematics/Algebra

We need to isolate $n$

$SAS_n = \dfrac{n}{2}\left[2a + d(n - 1)\right] \\[5ex]$ Swap. Let the LHS = RHS, and the RHS = LHS

$\dfrac{n}{2}\left[2a + d(n - 1)\right] = SAS_n \\[5ex]$ Remove the fraction. Find the $LCD$
$LCD = 2$
Multiply both sides by the $LCD$

$ 2 * \dfrac{n}{2}\left[2a + d(n - 1)\right] = 2 * SAS_n \\[5ex] n[2a + d(n - 1)] = 2SAS_n \\[3ex] n[2a + dn - d] = 2SAS_n \\[3ex] 2an + dn^2 - dn = 2SAS_n \\[3ex] dn^2 + 2an - dn = 2SAS_n \\[3ex] dn^2 + 2an - dn - 2SAS_n = 0 \\[3ex] (d)n^2 + n(2a - d) - 2SAS_n = 0 \\[3ex] This\:\: is\:\: a\:\: quadratic\:\: equation\:\: in\:\: n \\[3ex] Use\:\: the\:\: Quadratic\:\: Formula \\[3ex] ax^2 + bx + c = 0 \\[3ex] x = \dfrac{-b \pm \sqrt{b^2 - 4ac}}{2a} \\[5ex] Compare \\[3ex] a = d, b = 2a - d, c = -2SAS_n \\[3ex] n = \dfrac{-b \pm \sqrt{b^2 - 4ac}}{2a} \\[5ex] n = \dfrac{-(2a - d) \pm \sqrt{(2a - d)^2 - 4(d)(-2SAS_n)}}{2d} \\[5ex] n = \dfrac{-(2a - d) \pm \sqrt{(2a - d)^2 + 8d*SAS_n}}{2d} $
(17.) $ SAS_n = \dfrac{n}{2}[2a + d(n - 1)] \\[5ex] and \\[3ex] SAS_n = \dfrac{n}{2}(a + p) \\[5ex] $ for $n$


$ SAS_n = \dfrac{n}{2}[2a + d(n - 1)]...eqn.(1) \\[5ex] and \\[3ex] SAS_n = \dfrac{n}{2}(a + p)...eqn.(2) \\[5ex] $ for $n$
Sum of the $n$ terms of an Arithmetic Sequence: Mathematics/Algebra

We need to isolate $n$

$ SAS_n = SAS_n \\[3ex] \implies eqn.(1) = eqn.(2) \\[3ex] \dfrac{n}{2}[2a + d(n - 1)] = \dfrac{n}{2}(a + p) \\[5ex] Divide\:\: both\:\: sides\:\: by\:\: \dfrac{n}{2} \\[5ex] 2a + d(n - 1) = a + p \\[3ex] Subtract\:\: 2a \:\:from\:\: both\:\: sides \\[3ex] d(n - 1) = a + p - 2a \\[3ex] d(n - 1) = p - a \\[3ex] Divide\:\: both\:\: sides\:\: by\:\: d \\[3ex] n - 1 = \dfrac{p - a}{d} \\[5ex] Add\:\: 1 \:\:to\:\: both\:\: sides \\[3ex] n = \dfrac{p - a}{d} + 1 \\[5ex] n = \dfrac{p - a}{d} + \dfrac{d}{d} \\[5ex] n = \dfrac{p - a + d}{d} $
(18.) $AS_n = a + d(n - 1)$ for $n$
When given the last term, then $p = a + d(n - 1)$ for $n$


$p = a + d(n - 1)$ for $n$
The $n-th$ term of an Arithmetic Sequence: Mathematics/Algebra

We need to isolate $n$

$p = a + d(n - 1) \\[3ex]$ Swap. Let the LHS = RHS, and the RHS = LHS

$ a + d(n - 1) = p \\[3ex] Subtract\:\: a \:\:from\:\: both\:\: sides \\[3ex] d(n - 1) = p - a \\[3ex] Divide\:\: both\:\: sides\:\: by\:\: d \\[3ex] n - 1 = \dfrac{p - a}{d} \\[5ex] Add\:\: 1 \:\:to\:\: both\:\: sides \\[3ex] n = \dfrac{p - a}{d} + 1 \\[5ex] n = \dfrac{p - a}{d} + \dfrac{d}{d} \\[5ex] n = \dfrac{p - a + d}{d} $
(19.) ACT Each student's project in a history seminar is given a point score by the teacher and by each of the other students in the seminar.
A student's project grade, $g$ is determined by the formula $g = \dfrac{3t + s}{3 + n}$, where $t$ is the score the teacher gives, $s$ is the sum of the scores the students give, and $n$ is the number of students in the seminar.
What is $t$ in terms of $g$, $s$, and $n$?

$ F.\:\: t = g - n - s \\[3ex] G.\:\: t = gn + g - s \\[3ex] H.\:\: t = \dfrac{3gn - s}{9} \\[5ex] J.\:\: t = \dfrac{gn - s}{3} \\[3ex] K.\:\: t = \dfrac{3g + gn - s}{3} \\[5ex] $

We need to isolate $b$

$ g = \dfrac{3t + s}{3 + n} \\[5ex] Swap \\[3ex] \dfrac{3t + s}{3 + n} = g \\[5ex] Cross-multiply \\[3ex] 3t + s = g(3 + n) \\[3ex] Subtract \:\:s\:\: from\:\: both\:\: sides \\[3ex] 3t = g(3 + n) - s \\[3ex] 3t = 3g + 3n - s \\[3ex] Divide\:\: both\:\: sides\:\: by\:\: 3 \\[3ex] t = \dfrac{3g + gn - s}{3} $
(20.) ACT A formula for the area of a rhombus is $A = \dfrac{1}{2}d_1d_2$, where $d_1$ and $d_2$ are the lengths of the diagonals.
Which of the following is an expression for $d_2$?

$ F.\:\: \dfrac{2A}{d_1} \\[5ex] G.\:\: \dfrac{A}{2d_1} \\[5ex] H.\:\: \dfrac{Ad_1}{2} \\[5ex] J.\:\: 2(A - d_1) \\[3ex] K.\:\: A - \dfrac{d_1}{2} \\[3ex] $

$ A = \dfrac{1}{2}d_1d_2 \\[5ex] Swap \\[3ex] LCD = 2 \\[3ex] Multiply \:\:both\:\: sides\:\: by LCD \\[3ex] 2 * \dfrac{1}{2}d_1d_2 = 2 * A \\[5ex] d_1 * d_2 = 2A \\[3ex] Divide \:\:both\:\: sides\:\: by\:\: d_1 \\[3ex] d_2 = \dfrac{2A}{d_1} $




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(21.) WASSCE If $R = \dfrac{h}{2} + \dfrac{d^2}{8h}$, express $d$ in terms of $R$ and $h$

$ A.\:\: \sqrt{4h(2R - h)} \\[3ex] B.\:\: \sqrt{4h(R - h)} \\[3ex] C.\:\: \sqrt{4h(h - R)} \\[3ex] D.\:\: \sqrt{4h(h - 2R)} \\[3ex] $

We need to isolate $d$

$ R = \dfrac{h}{2} + \dfrac{d^2}{8h} \\[5ex] Swap \\[3ex] \dfrac{h}{2} + \dfrac{d^2}{8h} = R \\[5ex] Rearrange \\[3ex] \dfrac{d^2}{8h} + \dfrac{h}{2} = R \\[5ex] Subtract\:\:\dfrac{h}{2}\:\:from\:\:both\:\:sides \\[5ex] \dfrac{d^2}{8h} = R - \dfrac{h}{2} \\[5ex] Multiply\:\:both\:\:sides\:\:by\:\:8h \\[3ex] d^2 = 8h\left(R - \dfrac{h}{2}\right) \\[5ex] d^2 = 8hR - 4h^2 \\[3ex] d^2 = 4h(2R - h) \\[3ex] Take\:\:the\:\:Square\:\:Root\:\:of\:\:both\:\:sides \\[3ex] d = \sqrt{4h(2R - h)} $
(22.) JAMB If two graphs $y = px^2 + q$ and $y = 2x^2 - 1$ intersect at $x = 2$, find the value of $p$ in terms of $q$

$ A.\:\: \dfrac{q - 8}{7} \\[5ex] B.\:\: \dfrac{7 - q}{4} \\[5ex] C.\:\: \dfrac{8 - q}{2} \\[5ex] D.\:\: \dfrac{7 + q}{8} \\[5ex] $

The two graphs intersect at $x = 2$ means that they are equal at $x = 2$
Then we should isolate $p$

$ y = px^2 + q \\[3ex] When\:\:x = 2 \\[3ex] y = p(2)^2 + q \\[3ex] y = 4p + q \\[3ex] y = 2x^2 - 1 \\[3ex] When\:\:x = 2 \\[3ex] y = 2(2)^2 - 1 \\[3ex] y = 8 - 1 \\[3ex] y = 7 \\[3ex] y = y \\[3ex] \therefore 4p + q = 7 \\[3ex] Subtract\:\:q\:\:from\:\:both\:\:sides \\[3ex] 4p = 7 - q \\[3ex] Divide\:\:both\:\:sides\:\:by\:\:4 \\[3ex] p = \dfrac{7 - q}{4} $
(23.) CSEC (a) (i) Using the formula $t = \sqrt{\dfrac{5m}{12n}}$ calculate the value of $t$ when $m = 20$ and $n = 48$

(ii) Express $m$ as subject of the formula in (a) (i) above


$ (i) \\[3ex] t = \sqrt{\dfrac{5m}{12n}} \\[5ex] m = 20,\:\: n = 48 \\[3ex] t = \sqrt{\dfrac{5(20)}{12(48)}} \\[5ex] t = \sqrt{\dfrac{25}{144}} \\[5ex] t = \dfrac{5}{12} \\[5ex] (ii) \\[3ex] We\:\:need\:\:to\:\:isolate\:\:m \\[3ex] t = \sqrt{\dfrac{5m}{12n}} \\[5ex] Swap \\[3ex] \sqrt{\dfrac{5m}{12n}} = t \\[5ex] Square\:\:both\:\:sides \\[3ex] \dfrac{5m}{12n} = t^2 \\[5ex] Multiply\:\:both\:\:sides\:\:by\:\:12n \\[3ex] 5m = 12nt^2 \\[3ex] Divide\:\:both\:\:sides\:\:by\:\:5 \\[3ex] m = \dfrac{12nt^2}{5} $
(24.) CSEC The formula for the volume of a cylinder is given as $V = \pi r^2h$
Make $r$ the subject of the formula


We need to isolate $r$

$ V = \pi r^2 h \\[3ex] Swap \\[3ex] \pi r^2 h = V \\[3ex] Divide\:\:both\:\:sides\:\:by\:\:\pi h \\[3ex] r^2 = \dfrac{V}{\pi h} \\[5ex] Take\:\:the\:\:Square\:\:Root\:\:of\:\:both\:\:sides \\[3ex] r = \sqrt{\dfrac{V}{\pi h}} $
(25.) JAMB Make $r$ the subject of the formula $\dfrac{x}{r + a} = \dfrac{a}{r}$

$ A.\:\: \dfrac{a^2}{x + a} \\[5ex] B.\:\: \dfrac{a}{x - a} \\[5ex] C.\:\: \dfrac{a}{x + a} \\[5ex] D.\:\: \dfrac{a^2}{x - a} \\[5ex] $

We need to isolate $r$

$ \dfrac{x}{r + a} = \dfrac{a}{r} \\[5ex] Cross\:\:Multiply \\[3ex] xr = a(r + a) \\[3ex] xr = ar + a^2 \\[3ex] Collect\:\:like\:\:terms\:\:in\:\:r \\[3ex] xr - ar = a^2 \\[3ex] Factor\:\:by\:\:GCF \\[3ex] GCF = r \\[3ex] r(x - a) = a^2 \\[3ex] Divide\:\:both\:\:sides\:\:by\:\:(x - a) \\[3ex] r = \dfrac{a^2}{x - a} $
(26.) ACT Whenever $a^2b = 1$ for positive values of $a$ and $b$, which of the following equations gives $a$ in terms of $b$?

$ F.\:\: a = b^2 \\[3ex] G.\:\: a = \sqrt{b} \\[3ex] H.\:\: a = \dfrac{1}{b} \\[5ex] J.\:\: a = \dfrac{1}{b^2} \\[5ex] K.\:\: a = \dfrac{1}{\sqrt{b}} \\[5ex] $

We need to isolate $a$

$ a^2b = 1 \\[3ex] Divide\:\:both\:\:sides\:\:by\:\:b \\[3ex] a^2 = \dfrac{1}{b} \\[5ex] Take\:\:the\:\:Square\:\:Root\:\:of\:\:both\:\:sides \\[3ex] a = \sqrt{\dfrac{1}{b}} \\[5ex] a = \dfrac{\sqrt{1}}{\sqrt{b}} \\[5ex] a = \dfrac{1}{\sqrt{b}} $
(27.) ACT The value of a used car can be modeled by the formula $V = V_o(1 - r)^t$, where $V_o$ is the car's purchase price, in dollars; $r$ is the car's constant annual rate of decrease in value, expressed as a decimal; and $V$ is the car's dollar value at the end of $t$ years.
Which of the following equations shows the formula solved for $r$?

$ A.\:\: r = 1 - \sqrt[t]{\dfrac{V}{V_o}} \\[5ex] B.\:\: r = 1 + \sqrt[t]{\dfrac{V}{V_o}} \\[5ex] C.\:\: r = \sqrt[t]{\dfrac{V}{V_o}} - 1 \\[5ex] D.\:\: r = 1 - t\log\left(\dfrac{V}{V_o}\right) \\[5ex] E.\:\: r = t\log\left(\dfrac{V}{V_o}\right) - 1 \\[5ex] $

We need to isolate $r$

$ V = V_o(1 - r)^t \\[3ex] Swap \\[3ex] V_o(1 - r)^t = V \\[3ex] Divide\:\: both\:\: sides\:\: by\:\: V_o \\[3ex] (1 - r)^t = \dfrac{V}{V_o} \\[5ex] Take\:\:the\:\:t-th\:\:Root\:\:of\:\:both\:\:sides \\[3ex] OR \\[3ex] Multiply\:\:both\:\:exponents\:\:by\:\:\dfrac{1}{t} \\[5ex] 1 - r = \left(\dfrac{V}{V_o}\right)^{\dfrac{1}{t}} \\[7ex] 1 - r = \sqrt[t]{\dfrac{V}{V_o}} \\[5ex] Move\:\:and\:\:Transfer \\[3ex] 1 - \sqrt[t]{\dfrac{V}{V_o}} = r \\[5ex] Swap \\[3ex] r = 1 - \sqrt[t]{\dfrac{V}{V_o}} $
(28.) JAMB The weight $W$ kg of a metal bar varies jointly as its length $L$ metres and the square of its diameter $d$ metres.
If $W = 140$ when $d = 4\dfrac{2}{3}$ and $L = 54$, find $d$ in terms of $W$ and $L$

$ A.\:\: \sqrt{\dfrac{42W}{5L}} \\[5ex] B.\:\: \sqrt{\dfrac{5L}{42W}} \\[5ex] C.\:\: \dfrac{42W}{5L} \\[5ex] D.\:\: \dfrac{5L}{42W} \\[5ex] $

$ W \propto Ld^2...Joint\:\:Variation \\[3ex] W = kLd^2 ...Equation\:\:where\:\:k\:\:is\:\:a\:\:contant \\[3ex] We\:\:need\:\:to\:\:find\:\:k \\[3ex] Swap \\[3ex] kLd^2 = W \\[3ex] Divide\:\:both\:\:sides\:\:by\:\:Ld^2 \\[3ex] k = \dfrac{W}{Ld^2} \\[5ex] W = 140,\:\:d = 4\dfrac{2}{3} = \dfrac{14}{3},\:\:L = 54 \\[5ex] Ld^2 = 54 * \dfrac{14}{3} * \dfrac{14}{3} = 6 * 14 * 14 \\[5ex] \dfrac{W}{Ld^2} = \dfrac{140}{6 * 14 * 14} = \dfrac{10}{6 * 14} = \dfrac{5}{3 * 14} = \dfrac{5}{42} \\[5ex] \rightarrow k = \dfrac{5}{42} \\[5ex] W = kLd^2 \\[3ex] We\:\:need\:\:to\:\:isolate\:\:d \\[3ex] Swap \\[3ex] kLd^2 = W \\[3ex] Divide\:\:both\:\:sides\:\:by\:\:kL \\[3ex] d^2 = \dfrac{W}{kL} \\[5ex] kL = \dfrac{5}{42}L = \dfrac{5L}{42} \\[5ex] \dfrac{W}{kL} = W \div kL = W \div \dfrac{5L}{42} = W * \dfrac{42}{5L} = \dfrac{42W}{5L} \\[5ex] \rightarrow d^2 = \dfrac{42W}{5L} \\[5ex] Take\:\:the\:\:Square\:\:Root\:\:of\:\:both\:\:sides \\[3ex] d = \sqrt{\dfrac{42W}{5L}} $
(29.) CSEC (a) Given the formula $s = \dfrac{1}{2}(u + v)t$, express $u$ in terms of $v$, $s$, and $t$


We need to isolate $u$

$ s = \dfrac{1}{2}(u + v)t \\[5ex] Swap \\[3ex] \dfrac{1}{2}(u + v)t = s \\[5ex] Multiply\:\:both\:\:sides\:\:by\:\:2 \\[3ex] (u + v)t = 2s \\[3ex] Divide\:\:both\:\:sides\:\:by\:\:t \\[3ex] u + v = \dfrac{2s}{t} \\[5ex] Subtract\:\:v\:\:from\:\:both\:\:sides \\[3ex] u = \dfrac{2s}{t} - v \\[5ex] This\:\:is\:\:the\:\:answer \\[3ex] But,\:\:you\:\:may\:\:simplify\:\:further \\[3ex] u = \dfrac{2s}{t} - \dfrac{vt}{t} \\[5ex] u = \dfrac{2s - vt}{t} $
(30.) CSEC (a) Make $t$ the subject of the formula

$ \dfrac{p}{2} = \sqrt{\dfrac{t + r}{g}} \\[5ex] $

We need to isolate $t$

$ \dfrac{p}{2} = \sqrt{\dfrac{t + r}{g}} \\[5ex] Swap \\[3ex] \sqrt{\dfrac{t + r}{g}} = \dfrac{p}{2} \\[5ex] Square\:\:both\:\:sides \\[3ex] \dfrac{t + r}{g} = \left(\dfrac{p}{2}\right)^2 \\[5ex] \dfrac{t + r}{g} = \dfrac{p^2}{4} \\[5ex] Multiply\:\:both\:\:sides\:\:by\:\:g \\[3ex] t + r = \dfrac{gp^2}{4} \\[5ex] Subtract\:\:r\:\:from\:\:both\:\:sides \\[3ex] t = \dfrac{gp^2}{4} - r $
(31.) WASSCE Given that $\log_{10}y = 1 + 3\log_{10}x$, express $y$ in terms of $x$

$ A.\:\: y = 10x^3 \\[3ex] B.\:\: y = 10x^{-3} \\[3ex] C.\:\: y = x^3 \\[3ex] D.\:\: y = x^{-3} \\[3ex] $

We need to isolate $y$

$ \log_{10}y = 1 + 3\log_{10}x \\[3ex] 1 = \log_{10}10 ...Law\:\:4...Log \\[3ex] 3\log_{10}x = \log_{10}x^3 ...Law\:\:5...Log \\[3ex] \rightarrow \log_{10}y = \log_{10}10 + \log_{10}x^3 \\[3ex] \log_{10}10 + \log_{10}x^3 = \log_{10}(10 * x^3) ...Law\:\:1...Log \\[3ex] \rightarrow \log_{10}y = \log_{10}(10x^3) \\[3ex] Cancel\:\: \log_{10} \\[3ex] y = 10x^3 $
(32.) JAMB If $P = \sqrt{\dfrac{rs^3}{t}}$, express $r$ in terms of $P$, $s$ and $t$

$ A.\:\: \dfrac{P^2t}{s^3} \\[5ex] B.\:\: \dfrac{P^3t}{s^3} \\[5ex] C.\:\: \dfrac{P^3t}{s^2} \\[5ex] D.\:\: \dfrac{Pt}{s^3} \\[5ex] $

We need to isolate $r$

$ P = \sqrt{\dfrac{rs^3}{t}} \\[5ex] Swap \\[3ex] \sqrt{\dfrac{rs^3}{t}} = P \\[5ex] Square\:\:both\:\:sides \\[3ex] \dfrac{rs^3}{t} = P^2 \\[5ex] Multiply\:\:both\:\:sides\:\:by\:\:t \\[3ex] rs^3 = P^2t \\[3ex] Divide\:\:both\:\:sides\:\:by\:\:s^3 \\[3ex] r = \dfrac{P^2t}{s^3} $
(33.) WASSCE If $p, q, r, s$ are all positive and $p = \sqrt{q^2 - \dfrac{r^2}{s^2}}$, make $s$ the subject of the relation.

$ A.\:\: s = \dfrac{r}{q^2 - p^2} \\[5ex] B.\:\: s = \dfrac{r}{p^2 - q^2} \\[5ex] C.\:\: s = \dfrac{r}{\sqrt{q^2 - p^2}} \\[5ex] D.\:\: s = \dfrac{\sqrt{p^2 - q^2}}{r} \\[5ex] $

We need to isolate $s$

$ p = \sqrt{q^2 - \dfrac{r^2}{s^2}} \\[5ex] Swap \\[3ex] \sqrt{q^2 - \dfrac{r^2}{s^2}} = p \\[5ex] Square\:\:both\:\:sides \\[3ex] q^2 - \dfrac{r^2}{s^2} = p^2 \\[5ex] Move\:\:around \\[3ex] q^2 - p^2 = \dfrac{r^2}{s^2} \\[5ex] Multiply\:\:both\:\:sides\:\:by\:\:s^2 \\[3ex] s^2(q^2 - p^2) = r^2 \\[3ex] Divide\:\:both\:\:sides\:\:by\:\:(q^2 - p^2) \\[3ex] s^2 = \dfrac{r^2}{q^2 - p^2} \\[5ex] Take\:\:the\:\:Square\:\:Root\:\:of\:\:both\:\:sides \\[3ex] s = \sqrt{\dfrac{r^2}{q^2 - p^2}} \\[5ex] s = \dfrac{\sqrt{r^2}}{\sqrt{q^2 - p^2}} \\[5ex] s = \dfrac{r}{\sqrt{q^2 - p^2}} $
(34.) WASSCE Make $x$ the subject of the relation $\dfrac{x}{q} - 2 = \dfrac{x}{p}$

$ A.\:\: x = \dfrac{2pq}{p + q} \\[5ex] B.\:\: x = \dfrac{pq}{p - q} \\[5ex] C.\:\: x = \dfrac{pq}{p + q} \\[5ex] D.\:\: x = \dfrac{2pq}{p - q} \\[5ex] $

We need to isolate $x$
We can solve this in at least two ways.
You may use any method you like.

$ \dfrac{x}{q} - 2 = \dfrac{x}{p} \\[5ex] \underline{First\:\:Method} \\[3ex] LCD = pq \\[3ex] Multiply\:\:each\:\:term\:\:by\:\:the\:\:LCD \\[3ex] pq\left(\dfrac{x}{q}\right) - pq(2) = pq\left(\dfrac{x}{p}\right) \\[5ex] px - 2pg = qx \\[3ex] Collect\:\:like\:\:terms\:\:in\:\:x \\[3ex] Rearrange\\[3ex] px - qx = 2pq \\[3ex] Factor\:\:the\:\:LHS\:\:by\:\:GCF \\[3ex] GCF = x \\[3ex] x(p - q) = 2pq \\[3ex] Divide\:\:both\:\:sides\:\:by\:\:(p - q) \\[3ex] x = \dfrac{2pq}{p - q} \\[5ex] \underline{Second\:\:Method} \\[3ex] \dfrac{x}{q} - 2 = \dfrac{x}{p} \\[5ex] Collect\:\:like\:\:terms\:\:and\:\:Move\:\:around \\[3ex] \dfrac{x}{q} - \dfrac{x}{p} = 2 \\[5ex] \dfrac{px}{pq} - \dfrac{qx}{pq} = 2 \\[5ex] \dfrac{px - qx}{pq} = 2 \\[5ex] Multiply\:\:both\:\:sides\:\:by\:\:pq \\[3ex] px - qx = 2pq \\[3ex] Factor\:\:the\:\:LHS\:\:by\:\:GCF \\[3ex] GCF = x \\[3ex] x(p - q) = 2pq \\[3ex] Divide\:\:both\:\:sides\:\:by\:\:(p - q) \\[3ex] x = \dfrac{2pq}{p - q} $
(35.) WASSCE Given that $r = \dfrac{xy}{2}$ and $x = \dfrac{v}{w}$, express $r$ in terms of $y$, $v$, and $w$

$ A.\:\: \dfrac{2vw}{y} \\[5ex] B.\:\: \dfrac{vw}{2y} \\[5ex] C.\:\: \dfrac{vwy}{2} \\[5ex] D.\:\: \dfrac{vy}{2w} \\[5ex] $

We need to isolate $r$

$ r = \dfrac{xy}{2} \\[5ex] r = xy \div 2 \\[3ex] x = \dfrac{v}{w} \\[5ex] xy = \dfrac{v}{w} * y = \dfrac{vy}{w} \\[5ex] \rightarrow r = \dfrac{vy}{w} \div 2 \\[3ex] r = \dfrac{vy}{w} * \dfrac{1}{2} \\[5ex] r = \dfrac{vy}{2w} $
(36.) JAMB Make $Q$ the subject of the formula if $P = \dfrac{M}{5}(X + Q) + 1$

$ A.\:\: \dfrac{5P - MX - 5}{M} \\[5ex] B.\:\: {5P - MX - 5}{M} \\[5ex] C.\:\: {5P + MX + 5}{M} \\[5ex] D.\:\: {5P + MX - 5}{M} \\[5ex] $

We need to isolate $Q$

$ P = \dfrac{M}{5}(X + Q) + 1 \\[5ex] Swap \\[3ex] \dfrac{M}{5}(X + Q) + 1 = P \\[5ex] Subtract\:\:1\:\:from\:\:both\:\:sides \\[3ex] \dfrac{M}{5}(X + Q) = P - 1 \\[5ex] Multiply\:\:both\:\:sides\:\:by\:\:5 \\[3ex] M(X + Q) = 5(P - 1) \\[3ex] Divide\:\:both\:\:sides\:\:by\:\:M \\[3ex] X + Q = \dfrac{5(P - 1)}{M} \\[5ex] Subtract\:\:X\:\:from\:\:both\:\:sides \\[3ex] Q = \dfrac{5(P - 1)}{M} - X \\[5ex] Q = \dfrac{5(P - 1)}{M} - \dfrac{MX}{M} \\[5ex] Q = \dfrac{5(P - 1) - MX}{M} \\[5ex] Q = \dfrac{5P - 5 - MX}{M} \\[5ex] Q = \dfrac{5P - MX - 5}{M} $
(37.) CSEC (a) Make $r$ the subject of EACH of the following formulae:

$ (i)\:\: r - h = rh \\[3ex] (ii)\:\: V = \pi r^2h \\[3ex] $

We need to isolate $r$

$ (i) \\[3ex] r - h = rh \\[3ex] Collect\:\:like\:\:terms\:\:in\:\:r\:\:and\:\:Move\:\:around \\[3ex] r - rh = h \\[3ex] Factor\:\:the\:\:LHS\:\:by\:\:GCF \\[3ex] GCF = r \\[3ex] r(1 - h) = h \\[3ex] Divide\:\:both\:\:sides\:\:by\:\:(1 - h) \\[3ex] r = \dfrac{h}{1 - h} \\[5ex] (ii) \\[3ex] V = \pi r^2h \\[3ex] Swap \\[3ex] \pi r^2h = V \\[3ex] Divide\:\:both\:\:sides\:\:by\:\: \pi h \\[3ex] r^2 = \dfrac{V}{\pi h} \\[5ex] Take\:\:the\:\:Square\:\:Root\:\:of\:\:both\:\:sides \\[3ex] r = \sqrt{\dfrac{V}{\pi h}} $
(38.) CSEC (a) (i) Make $x$ the subject of the formula

$ y = \dfrac{x}{5} + 3p \\[5ex] $

We need to isolate $x$

$ y = \dfrac{x}{5} + 3p \\[5ex] Swap \\[3ex] \dfrac{x}{5} + 3p = y \\[5ex] Subtract\:\:3p\:\:from\:\:both\:\:sides \\[3ex] \dfrac{x}{5} = y - 3p \\[5ex] Multiply\:\:both\:\:sides\:\:by\:\:5 \\[3ex] x = 5(y - 3p) $
(39.) JAMB Make $R$ the subject of the formula if $T = \dfrac{KR^2 + M}{3}$

$ A.\:\: \sqrt{\dfrac{3T + M}{K}} \\[5ex] B.\:\: \sqrt{\dfrac{3T - K}{M}} \\[5ex] C.\:\: \sqrt{\dfrac{3T + K}{M}} \\[5ex] D.\:\: \sqrt{\dfrac{3T - M}{K}} \\[5ex] $

We need to isolate $R$

$ T = \dfrac{KR^2 + M}{3} \\[5ex] Multiply\:\:by\:\:both\:\:by\:\:3 \\[3ex] 3T = KR^2 + M \\[3ex] Swap \\[3ex] KR^2 + M = 3T \\[3ex] Subtract\:\:M\:\:from\:\:both\:\:sides \\[3ex] KR^2 = 3T - M \\[3ex] Divide\:\:both\:\:sides\:\:by\:\:K \\[3ex] R^2 = \dfrac{3T - M}{K} \\[5ex] Take\:\:the\:\:Square\:\:Root\:\:of\:\:both\:\:sides \\[3ex] R = \sqrt{\dfrac{3T - M}{K}} $
(40.) WASSCE Given that $p^{\dfrac{1}{3}} = \dfrac{\sqrt[3]{q}}{r}$, make $q$ the subject of the equation.

$ A.\:\: q = p\sqrt{r} \\[3ex] B.\:\: q = p^3r \\[3ex] C.\:\: q = pr^3 \\[3ex] D.\:\: q = pr^{\dfrac{1}{3}} \\[3ex] $

We need to isolate $q$

$ p^{\dfrac{1}{3}} = \dfrac{\sqrt[3]{q}}{r} \\[5ex] Swap \\[3ex] \dfrac{\sqrt[3]{q}}{r} = p^{\dfrac{1}{3}} \\[5ex] Multiply\:\:both\:\:sides\:\:by\:\:r \\[3ex] \sqrt[3]{q} = rp^{\dfrac{1}{3}} \\[3ex] Cube\:\:both\:\:sides \\[3ex] q = \left(rp^{\dfrac{1}{3}}\right)^3 \\[3ex] q = r^3 * p \\[3ex] q = pr^3 $




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(41.) JAMB Make $n$ the subject of the formula if $w = \dfrac{v(2 + cn)}{1 - cn}$

$ A.\:\: \dfrac{1}{c}\left(\dfrac{w - 2v}{v + w}\right) \\[5ex] B.\:\: \dfrac{1}{c}\left(\dfrac{w - 2v}{v - w}\right) \\[5ex] C.\:\: \dfrac{1}{c}\left(\dfrac{w + 2v}{v - w}\right) \\[5ex] D.\:\: \dfrac{1}{c}\left(\dfrac{w + 2v}{v + w}\right) \\[5ex] $

We need to isolate $n$

$ w = \dfrac{v(2 + cn)}{1 - cn} \\[5ex] Multiply\:\:both\:\:sides\:\:by\:\:(1 - cn) \\[3ex] w(1 - cn) = v(2 + cn) \\[3ex] Expand \\[3ex] w - wcn = 2v + vcn \\[3ex] Collect\:\:like\:\:terms\:\:in\:\:n \\[3ex] Move\:\:around \\[3ex] w - 2v = vcn + wcn \\[3ex] Swap \\[3ex] vcn + wcn = w - 2v \\[3ex] Factor\:\:by\:\:GCF \\[3ex] GCF = cn \\[3ex] cn(v + w) = w - 2v \\[3ex] Divide\:\:both\:\:sides\:\:by\:\:c(v + w) \\[3ex] n = \dfrac{w - 2v}{c(v + w)} \\[5ex] \rightarrow n = \dfrac{1}{c}\left(\dfrac{w - 2v}{v + w}\right) $
(42.) WASSCE Make $m$ the subject of the equation $y = mx + c$

$ A.\:\: m = \dfrac{y - x}{c} \\[5ex] B.\:\: m = \dfrac{y - c}{x} \\[5ex] C.\:\: m = x(y - c) \\[3ex] D.\:\: m = x(y + c) \\[3ex] $

We need to isolate $m$

$ y = mx + c \\[3ex] Swap \\[3ex] mx + c = y \\[3ex] Subtract\:\:c\:\:from\:\:both\:\:sides \\[3ex] mx = y - c \\[3ex] Divide\:\:both\:\:sides\:\:by\:\:x \\[3ex] m = \dfrac{y - c}{x} $
(43.) CSEC (c) When a number, $x$ is multiplied by $2$, the result is squared to give a new number, $y$.
(i) Express $y$ in terms of $x$
(ii) Determine the two values of $x$ that satisfy the equation $y = x$ AND the equation derived in (c) (i)


$ x\:\:multiplied\:\:by\:\:2 = x * 2 = 2x \\[3ex] result\:\:is\:\:squared = (2x)^2 = 4x^2 \\[3ex] \rightarrow y = 4x^2 \\[3ex] (i) \\[3ex] y = 4x^2 \\[3ex] (ii) \\[3ex] y = x \\[3ex] y = 4x^2 \\[3ex] y = y \rightarrow x = 4x^2 \\[3ex] x = 4x^2 \\[3ex] 0 = 4x^2 - x \\[3ex] 4x^2 - x = 0 \\[3ex] x(4x - 1) = 0 \\[3ex] x = 0 \:\:OR\:\: 4x - 1 = 0 \\[3ex] 4x - 1 = 0 \\[3ex] 4x = 1 \\[3ex] x = \dfrac{1}{4} \\[5ex] x = 0 \:\:OR\:\: x = \dfrac{1}{4} $
(44.) CSEC (b) (i) Make $C$ the subject of the formula $F = \dfrac{9}{5}C + 32$

(ii) Given that $F = 113$, calculate the value of $C$


We need to isolate $C$

$ (i) \\[3ex] F = \dfrac{9}{5}C + 32 \\[5ex] Swap \\[3ex] \dfrac{9}{5}C + 32 = F \\[5ex] Subtract\:\:32\:\:from\:\:both\:\:sides \\[3ex] \dfrac{9}{5}C = F - 32 \\[5ex] Multiply\:\:both\:\:sides\:\:by\:\:5 \\[3ex] 9C = 5(F - 32) \\[3ex] Divide\:\:both\:\:sides\:\:by\:\:9 \\[3ex] C = \dfrac{5(F - 32)}{9} \\[5ex] (ii) \\[3ex] F = 113 \\[3ex] C = \dfrac{5(113 - 32)}{9} \\[5ex] C = \dfrac{5(81)}{9} \\[5ex] C = 5(9) \\[3ex] C = 45^\circ C $
(45.) CSEC (c) Adam, Imran, and Shakeel were playing a card game.
Adam scored $x$ points
Imran scored $3$ points fewer than Adam
Shakeel scored twice as many as many points as Imran
Together, they scored $39$ points.
(i) Write down, in terms of $x$, an expression for the number of points scored by Shakeel.
(ii) Write an equation which may be used to find the value of $x$


$ Adam's\:\:points = x \\[3ex] 3\:\:points\:\:fewer\:\:than\:\:x = x - 3 \\[3ex] \rightarrow Imran's\:\:points = x - 3 \\[3ex] twice\:\:as\:\:many\:\:points\:\:as\:\:(x - 3) = 2(x - 3) \\[3ex] \rightarrow Shakeel's\:\:points = 2(x - 3) \\[3ex] (i) \\[3ex] Shakeel's\:\:points = 2(x - 3) \\[3ex] (ii) \\[3ex] \therefore x + (x - 3) + 2(x - 3) = 39 \\[3ex] x + x - 3 + 2x - 6 = 39 \\[3ex] 4x - 9 = 39 \\[3ex] 4x = 39 + 9 \\[3ex] 4x = 48 \\[3ex] x = \dfrac{48}{4} \\[5ex] x = 12 $
(46.) CSEC (c) A formula is given as $d = \sqrt{\dfrac{4h}{5}}$

(i) Determine the value of $d$ when $h = 29$. Give your answer correct to $3$ significant figures
(ii) Make $h$ the subject of the formula


$ (i) \\[3ex] h = 29 \\[3ex] d = \sqrt{\dfrac{4h}{5}} \\[5ex] d = \sqrt{\dfrac{4(29)}{5}} \\[5ex] d = \sqrt{\dfrac{116}{5}} \\[5ex] d = \sqrt{23.2} \\[3ex] d = 4.816637832 \\[3ex] d \approx 4.82 \\[3ex] (ii) \\[3ex] d = \sqrt{\dfrac{4h}{5}} \\[5ex] We\:\:need\:\:to\:\:isolate\:\:h \\[3ex] Swap \\[3ex] \sqrt{\dfrac{4h}{5}} = d \\[5ex] Square\:\:both\:\:sides \\[3ex] \dfrac{4h}{5} = d^2 \\[5ex] Multiply\:\:both\:\:sides\:\:by\:\:5 \\[3ex] 4h = 5d^2 \\[3ex] Divide\:\:both\:\:sides\:\:by\:\:4 \\[3ex] h = \dfrac{5h^2}{4} $
(47.) CSEC (c) The quantities $F, m, u, v\:\:and\:\:t$ are related according to the formula $F = \dfrac{m(v - u)}{t}$

(i) Find the value of $F$ when $m = 3, u = -1, v = 2\:\:and\:\: t = 1$

(ii) Make $v$ the subject of the formula


$ (i) \\[3ex] F = \dfrac{m(v - u)}{t} \\[5ex] m = 3, u = -1, v = 2, t = 1 \\[3ex] F = \dfrac{3(2 - -1)}{1} \\[5ex] F = 3(2 + 1) \\[3ex] F = 3(3) \\[3ex] F = 6 \\[3ex] (ii) \\[3ex] F = \dfrac{m(v - u)}{t} \\[5ex] We\:\:need\:\:to\:\:isolate\:\:v \\[3ex] Swap \\[3ex] \dfrac{m(v - u)}{t} = F \\[5ex] Multiply\:\:both\:\:sides\:\:by\:\:t \\[3ex] m(v - u) = Ft \\[3ex] Divide\:\:both\:\:sides\:\:by\:\:m \\[3ex] v - u = \dfrac{Ft}{m} \\[5ex] Add\:\:u\:\:to\:\:both\:\:sides \\[3ex] v = \dfrac{Ft}{m} + u \\[5ex] You\:\:may\:\:simplify\:\:further \\[3ex] v = \dfrac{Ft}{m} + \dfrac{mu}{m} \\[5ex] v = \dfrac{Ft + mu}{m} $
(48.) ACT Given that $c = 10b^3 + 50$, which of the following is an expression for $b$ in terms of $c$?

$ F.\:\: \left(\dfrac{c}{10} - 5\right)^{\dfrac{1}{3}} \\[5ex] G.\:\: \left(\dfrac{c}{10} + 5\right)^{\dfrac{1}{3}} \\[5ex] H.\:\: \dfrac{1}{10}(c - 50)^{\dfrac{1}{3}} \\[5ex] J.\:\: c^3 + 5 \\[3ex] K.\:\: 10c^3 + 50 \\[3ex] $

This means that we should isolate $b$

$ c = 10b^3 + 50 \\[3ex] 10b^3 + 50 = c \\[3ex] 10b^3 = c - 50 \\[3ex] b^3 = \dfrac{c - 50}{10} \\[5ex] b^3 = \dfrac{c}{10} - \dfrac{50}{10} \\[5ex] b^3 = \dfrac{c}{10} - 5 \\[5ex] $ To isolate $b$, multiply the exponents of both sides by $\dfrac{1}{3}$

$ (b^3)^{\dfrac{1}{3}} = \left(\dfrac{c}{10} - 5\right)^{\dfrac{1}{3}} \\[5ex] b = \left(\dfrac{c}{10} - 5\right)^{\dfrac{1}{3}} $
(49.) JAMB If $S = \sqrt{t^2 - 4t + 4}$ find $t$ in terms of $S$

$ A.\:\: S + 2 \\[3ex] B.\:\: S - 2 \\[3ex] C.\:\: S^2 + 2 \\[3ex] D.\:\: S^2 - 2 \\[3ex] $

We need to isolate $t$

$ S = \sqrt{t^2 - 4t + 4} \\[3ex] Square\:\:both\:\:sides \\[3ex] S^2 = t^2 - 4t + 4 \\[3ex] Swap \\[3ex] t^2 - 4t + 4 = S^2 \\[3ex] Factor\:\:the\:\:LHS \\[3ex] (t - 2)(t - 2) = S^2 \\[3ex] (t - 2)^2 = S^2 \\[3ex] Take\:\:the\:\:Square\:\:Root\:\:of\:\:both\:\:sides \\[3ex] t - 2 = S \\[3ex] Add\:\:2\:\:to\:\:both\:\:sides \\[3ex] t = S + 2 $
(50.) ACT Whenever $3(a + 6b) - c = 0$, which of the following expressions is equal to $a$

$ F.\:\: \dfrac{c}{3} - 6b \\[5ex] G.\:\: \dfrac{c}{3} - 2b \\[5ex] H.\:\: 2b - \dfrac{c}{3} \\[5ex] J.\:\: c - 6b \\[3ex] K.\:\: 3c - 6b \\[3ex] $

We need to isolate $a$

$ 3(a + 6b) - c = 0 \\[3ex] Add\:\:c\:\:to\:\:both\:\:sides \\[3ex] 3(a + 6b) = 0 + c \\[3ex] 3(a + 6b) = c \\[3ex] Divide\:\:both\:\:sides\:\:by\:\:3 \\[3ex] a + 6b = \dfrac{c}{3} \\[5ex] Subtract\:\:6b\:\:from\:\:both\:\:sides \\[3ex] a = \dfrac{c}{3} - 6b $
(51.) JAMB Make $T$ the subject of the equation $\dfrac{av}{1 - v} = \sqrt[3]{\dfrac{2v + T}{a + 2T}}$

$ A.\:\: T = \dfrac{3av}{1 - v} \\[5ex] B.\:\: T = \dfrac{1 + v}{2a^2v^3} \\[5ex] C.\:\: T = \dfrac{2v(1 - v)^3 - a^4v^3}{2a^3v^3 + (1 - v)^2} \\[5ex] D.\:\: T = \dfrac{2v(1 - v)^3 - a^4v^3}{2a^3v^3 - (1 - v)^3} \\[5ex] $

We need to isolate $T$

$ \dfrac{av}{1 - v} = \sqrt[3]{\dfrac{2v + T}{a + 2T}} \\[5ex] Cube\:\:both\:\:sides \\[3ex] \left(\dfrac{av}{1 - v}\right)^3 = \dfrac{2v + T}{a + 2T} \\[5ex] \dfrac{(av)^3}{(1 - v)^3} = \dfrac{2v + T}{a + 2T} \\[5ex] (av)^3 = a^3v^3 \\[3ex] (1 - v)^3 = 1^3 + 3(1)^2(-v) + 3(1)(-v)^2 + (-v)^3 ...Pascal's\:\:Triangle \\[3ex] (1 - v)^3 = 1 + 3(1)(-v) + 3(1)(v^2) + -v^3 \\[3ex] (1 - v)^3 = 1 - 3v + 3v^2 - v^3 \\[3ex] \rightarrow \dfrac{a^3v^3}{1 - 3v + 3v^2 - v^3} = \dfrac{2v + T}{a + 2T} \\[5ex] Cross\:\:Multiply \\[3ex] a^3v^3(a + 2T) = (2v + T)(1 - 3v + 3v^2 - v^3) \\[3ex] a^4v^3 + 2a^3v^3T = 2v - 6v^2 + 6v^3 - 2v^4 + T - 3vT + 3v^2T - v^3T \\[3ex] Collect\:\:like\:\:terms\:\:in\:\:T \\[3ex] Move\:\:things\:\:around\:\: \\[3ex] 2a^3v^3T - T + 3vT - 3v^2T + v^3T = 2v - 6v^2 + 6v^3 - 2v^4 - a^4v^3 \\[3ex] T(2a^3v^3 - 1 + 3v - 3v^2 + v^3) = 2v - 6v^2 + 6v^3 - 2v^4 - a^4v^3 \\[3ex] Divide\:\:both\:\:sides\:\:by\:\:(2a^2v^2 - 1 + 2v - v^2) \\[3ex] T = \dfrac{2v - 6v^2 + 6v^3 - 2v^4 - a^4v^3}{2a^3v^3 - 1 + 3v - 3v^2 + v^3} \\[5ex] We\:\:need\:\:to\:\:get\:\:the\:\:answer\:\:in\:\:the\:\:option \\[3ex] So\:\:we\:\:need\:\:to\:\:simplify\:\:further \\[3ex] \underline{Numerator} \\[3ex] 2v - 6v^2 + 6v^3 - 2v^4 - a^4v^3 = 2v(1 - 3v + 3v^2 - v^3) - a^4v^3 \\[3ex] But\:\: (1 - v)^3 = 1 - 3v + 3v^2 - v^3 \\[3ex] So\:\: 1 - 3v + 3v^2 - v^3 = (1 - v)^3 \\[3ex] \rightarrow 2v - 6v^2 + 6v^3 - 2v^4 - a^4v^3 = 2v(1 - v)^3 - a^4v^3 \\[3ex] \underline{Denominator} \\[3ex] 2a^3v^3 - 1 + 3v - 3v^2 + v^3 = 2a^3v^3 - (1 - 3v + 3v^2 - v^3) \\[3ex] But\:\: 1 - 3v + 3v^2 - v^3 = (1 - v)^3 \\[3ex] \rightarrow 2a^3v^3 - 1 + 3v - 3v^2 + v^3 = 2a^3v^3 - (1 - v)^3 \\[3ex] \therefore T = \dfrac{2v(1 - v)^3 - a^4v^3}{2a^3v^3 - (1 - v)^3} $
(52.) ACT If $3x + a = 9$, then, in terms of $a$, $x = ?$

$ A.\:\: \dfrac{a}{6} \\[5ex] B.\:\: 3a \\[3ex] C.\:\: 3 - a \\[3ex] D.\:\: \dfrac{9 - a}{3} \\[5ex] E.\:\: \dfrac{9 + a}{3} \\[5ex] $

We need to isolate $x$

$ 3x + a = 9 \\[3ex] Subtract\:\:a\:\:from\:\:both\:\:sides \\[3ex] 3x = 9 - a \\[3ex] Divide\:\:both\:\:sides\:\:by\:\:3 \\[3ex] x = \dfrac{9 - a}{3} $
(53.) JAMB If $pq + 1 = q^2$ and $t = \dfrac{1}{p} - \dfrac{1}{pq}$, express $t$ in terms of $q$

$ A.\:\: \dfrac{1}{p - q} \\[5ex] B.\:\: \dfrac{1}{q - 1} \\[5ex] C.\:\: \dfrac{1}{q + 1} \\[5ex] D.\:\: 1 + q \\[3ex] E.\:\: \dfrac{1}{1 - q} \\[5ex] $

We need to isolate $t$ and express it in terms of $q$

$ t = \dfrac{1}{p} - \dfrac{1}{pq} ...eqn(1) \\[5ex] pq + 1 = q^2 \\[3ex] \rightarrow pq = q^2 - 1 \\[3ex] \dfrac{1}{pq} = \dfrac{1}{q^2 - 1} \\[5ex] \rightarrow p = \dfrac{q^2 - 1}{q} \\[5ex] \dfrac{1}{p} = \dfrac{q}{q^2 - 1} \\[5ex] Substitute\:\:in\:\:eqn.(1) \\[3ex] \rightarrow t = \dfrac{q}{q^2 - 1} - \dfrac{1}{q^2 - 1} \\[5ex] t = \dfrac{q - 1}{q^2 - 1} \\[5ex] q^2 - 1 = (q + 1)(q - 1)...Difference\:\:of\:\:Two\:\:Squares \\[3ex] \rightarrow t = \dfrac{q - 1}{(q + 1)(q - 1)} \\[5ex] t = \dfrac{1}{q + 1} $
(54.) ACT For the equation $5y + n = m$, which of the following expressions gives $y$ in terms of $m$ and $n$?

$ F.\:\: \dfrac{m - n}{5} \\[5ex] G.\:\: \dfrac{m - 5}{n} \\[5ex] H.\:\: \dfrac{m + n}{5} \\[5ex] J.\:\: \dfrac{n - m}{5} \\[5ex] K.\:\: m - n - 5 \\[3ex] $

We need to isolate $y$

$ 5y + n = m \\[3ex] Subtract\:\:n\:\:from\:\:both\:\:sides \\[3ex] 5y = m - n \\[3ex] Divide\:\:both\:\:sides\:\:by\:\:5 \\[3ex] y = \dfrac{m - n}{5} $
(55.) JAMB Write $h$ in terms of $a, b, c, d$ if $a = \dfrac{b(1 - ch)}{1 - dh}$

$ A.\:\: h = \dfrac{a - b}{ad - bc} \\[5ex] B.\:\: h = \dfrac{a + b}{ad - bc} \\[5ex] C.\:\: h = \dfrac{ad - bc}{a - b} \\[5ex] D.\:\: h = \dfrac{1 - b}{d - bc} \\[5ex] E.\:\: h = \dfrac{b - a}{ad - bc} \\[5ex] $

We need to isolate $h$

$ a = \dfrac{b(1 - ch)}{1 - dh} \\[5ex] Cross\:\:Multiply \\[3ex] a(1 - dh) = b(1 - ch) \\[3ex] a - adh = b - bch \\[3ex] Collect\:\:like\:\:terms\:\:in\:\:h \\[3ex] Move\:\:terms \\[3ex] a - b = -bch + adh \\[3ex] a - b = adh - bch \\[3ex] Swap \\[3ex] adh - bch = a - b \\[3ex] h(ad - bc) = a - b \\[3ex] Divide\:\:both\:\:sides\:\:by\:\:(ad - bc) \\[3ex] h = \dfrac{a - b}{ad - bc} $
(56.) ACT When $bd - k = y$ and $d \ne 0$, $b = ?$

$ F.\:\: \dfrac{y + k}{d} \\[5ex] G.\:\: \dfrac{y - k}{d} \\[5ex] H.\:\: k - y - d \\[3ex] J.\:\: y - k + d \\[3ex] K.\:\: y + k - d \\[3ex] $

We need to isolate $b$

$ bd - k = y \\[3ex] Add\:\:k\:\:to\:\:both\:\:sides \\[3ex] bd = y + k \\[3ex] Divide\:\:both\:\:sides\:\:by\:\:d \\[3ex] b = \dfrac{y + k}{d} $
(57.) JAMB Make $u$ the subject of the formula $s = \sqrt{\dfrac{6}{u} - \dfrac{w}{2}}$

$ A.\:\: u = \dfrac{12}{2s^2} \\[5ex] B.\:\: u = \dfrac{12}{2s + w} \\[5ex] C.\:\: u = \dfrac{12}{2s^2 + w} \\[5ex] D.\:\: u = \dfrac{12}{2s^2} + w \\[5ex] $

We need to isolate $u$

$ s = \sqrt{\dfrac{6}{u} - \dfrac{w}{2}} \\[5ex] Swap \\[3ex] \sqrt{\dfrac{6}{u} - \dfrac{w}{2}} = s \\[5ex] Square\:\:both\:\:sides \\[3ex] \dfrac{6}{u} - \dfrac{w}{2} = s^2 \\[5ex] Add\:\:\dfrac{w}{2}\:\:to\:\:both\:\:sides \\[5ex] \dfrac{6}{u} = s^2 + \dfrac{w}{2} \\[5ex] \dfrac{6}{u} = \dfrac{2s^2}{2} + \dfrac{w}{2} \\[5ex] \dfrac{6}{u} = \dfrac{2s^2 + w}{2} \\[5ex] Take\:\:the\:\:Reciprocal\:\:of\:\:both\:\:sides \\[3ex] \dfrac{u}{6} = \dfrac{2}{2s^2 + w} \\[5ex] Multiply\:\:both\:\:sides\:\:by\:\:6 \\[3ex] u = 6 * \dfrac{2}{2s^2 + w} \\[5ex] u = \dfrac{12}{2s^2 + w} $
(58.) ACT For all nonzero values of $a, b,\:\:and\:\:c$, which of the following is the solution for $x$ pf the equation $ax + b = c$?

$ F.\:\: \dfrac{c}{ab} \\[5ex] G.\:\: \dfrac{c}{a} - b \\[5ex] H.\:\: \dfrac{b - c}{a} \\[5ex] J.\:\: \dfrac{b + c}{a} \\[5ex] K.\:\: \dfrac{c - b}{a} \\[5ex] $

We need to isolate $x$

$ ax + b = c \\[3ex] Subtract\:\:b\:\:from\:\:both\:\:sides \\[3ex] ax = c - b \\[3ex] Divide\:\:both\:\:sides\:\:by\:\:a \\[3ex] x = \dfrac{c - b}{a} $
(59.) JAMB Make $y$ the subject of the formula $z = x^2 + \dfrac{1}{y^3}$

$ A.\:\: y = \dfrac{1}{(z - x^2)^3} \\[5ex] B.\:\: y = \dfrac{1}{(z + x^3)^{\dfrac{1}{3}}} \\[7ex] C.\:\: y = \dfrac{1}{(z - x^2)^{\dfrac{1}{3}}} \\[7ex] D.\:\: y = \dfrac{1}{\sqrt[3]{z} - \sqrt[3]{x^2}} \\[5ex] $

We need to isolate $y$

$ z = x^2 + \dfrac{1}{y^3} \\[5ex] Subtract\:\:x^2\:\:from\:\:both\:\:sides \\[3ex] z - x^2 = \dfrac{1}{y^3} \\[5ex] Multiply\:\:both\:\:sides\:\:by\:\:y^3 \\[3ex] y^3(z - x^2) = 1 \\[3ex] Divide\:\:both\:\:sides\:\:by\:\:(z - x^2) \\[3ex] y^3 = \dfrac{1}{z - x^2} \\[5ex] Take\:\:the\:\:Cube\:\:Root\:\:of\:\:both\:\:sides \\[3ex] y = \sqrt[3]{\dfrac{1}{z - x^2}} \\[5ex] y = \dfrac{\sqrt[3]{1}}{\sqrt[3]{z - x^2}} \\[5ex] y = \dfrac{1}{\sqrt[3]{z - x^2}} \\[5ex] y = \dfrac{1}{(z - x^2)^{\dfrac{1}{3}}} $
(60.) ACT Which of the following is an equivalent expression for $r$ in terms of $S$ and $t$ whenever $r, S,\:\:and\:\: t$ are all distinct and $S = \dfrac{rt - 3}{r - t}$

$ A.\:\: \dfrac{St - 3}{S - t} \\[5ex] B.\:\: \dfrac{S - 3}{S - 1} \\[5ex] C.\:\: \dfrac{S - t}{S - 3} \\[5ex] D.\:\: \dfrac{St - 3}{S + t} \\[5ex] E.\:\: \dfrac{3}{t - S} \\[5ex] $

We need to isolate $r$

$ S = \dfrac{rt - 3}{r - t} \\[5ex] Multiply\:\:both\:\:sides\:\:by\:\:(r - t) \\[3ex] S(r - t) = rt - 3 \\[3ex] Sr - St = rt - 3 \\[3ex] Collect\:\:like\:\:terms\:\:in\:\:r \\[3ex] Move\:\:terms \\[3ex] Sr - rt = -3 + St \\[3ex] Sr - rt = St - 3 \\[3ex] Factor\:\:the\:\:LHS\:\:GCF \\[3ex] GCF = r \\[3ex] r(S - t) = St - 3 \\[3ex] Divide\:\:both\:\:sides\:\:by\:\:(S - t) \\[3ex] r = \dfrac{St - 3}{S - t} $




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(61.) JAMB If $\dfrac{x}{a + 1} + \dfrac{y}{b} = 1$, make $y$ the subject of the relation

$ A.\:\: \dfrac{b(a + 1 - x)}{a + 1} \\[5ex] B.\:\: \dfrac{a + 1}{b(a - x + 1)} \\[5ex] C.\:\: \dfrac{a(b - x + 1)}{b + 1} \\[5ex] D.\:\: \dfrac{b}{a(b - x + 1)} \\[5ex] $

We need to isolate $y$

$ \dfrac{x}{a + 1} + \dfrac{y}{b} = 1 \\[5ex] Subtract\:\:\dfrac{x}{a + 1}\:\:from\:\:both\:\:sides \\[5ex] \dfrac{y}{b} = 1 - \dfrac{x}{a + 1} \\[5ex] \dfrac{y}{b} = \dfrac{a + 1}{a + 1} - \dfrac{x}{a + 1} \\[5ex] \dfrac{y}{b} = \dfrac{(a+ 1) - x}{a + 1} \\[5ex] \dfrac{y}{b} = \dfrac{a + 1 - x}{a + 1} \\[5ex] Multiply\:\:both\:\:sides\:\:by\:\:b \\[3ex] y = \dfrac{b(a + 1 - x)}{a + 1} $
(62.) JAMB Given $M = N\sqrt{\dfrac{SL}{T}}$, make $T$ the subject of the formula.

$ A.\:\: \dfrac{NSL}{M} \\[5ex] B.\:\: \dfrac{N^2SL}{M^2} \\[5ex] C.\:\: \dfrac{N^2SL}{M} \\[5ex] D.\:\: \dfrac{NSL}{M^2} \\[5ex] $

We need to isolate $T$

$ M = N\sqrt{\dfrac{SL}{T}} \\[5ex] Divide\:\:both\:\:sides\:\:by\:\:N \\[3ex] \dfrac{M}{N} = \sqrt{\dfrac{SL}{T}} \\[5ex] Square\:\:both\:\:sides \\[3ex] \left(\dfrac{M}{N}\right)^2 = \dfrac{SL}{T} \\[5ex] \dfrac{M^2}{N^2} = \dfrac{SL}{T} \\[5ex] Cross\:\:Multiply \\[3ex] M^2T = N^2SL \\[3ex] Divide\:\:both\:\:sides\:\:by\:\:M^2 \\[3ex] T = \dfrac{N^2SL}{M^2} $
(63.) JAMB If $P = \left[\dfrac{Q(R - T)}{15}\right]^{\dfrac{1}{3}}$, make $T$ the subject of the relation

$ A.\:\: T = \dfrac{R + P^3}{15Q} \\[5ex] B.\:\: T = \dfrac{R - 15P^3}{Q} \\[5ex] C.\:\: T = R - \dfrac{15P^3}{Q} \\[5ex] D.\:\: T = \dfrac{15R + Q}{P^3} \\[5ex] $

We need to isolate $T$

$ P = \left[\dfrac{Q(R - T)}{15}\right]^{\dfrac{1}{3}} \\[5ex] Swap \\[3ex] \left[\dfrac{Q(R - T)}{15}\right]^{\dfrac{1}{3}} = P \\[3ex] Cube\:\:both\:\:sides \\[3ex] \dfrac{Q(R - T)}{15} = P^3 \\[3ex] Multiply\:\:both\:\:sides\:\:by\:\:15 \\[3ex] Q(R - T) = 15P^3 \\[3ex] Divide\:\:both\:\:sides\:\:by\:\:Q \\[3ex] R - T = \dfrac{15P^3}{Q} \\[5ex] Move\:\:around \\[3ex] R - \dfrac{15P^3}{Q} = T \\[5ex] Swap \\[3ex] T = R - \dfrac{15P^3}{Q} $
(64.) JAMB For $p = s + \dfrac{sm^2}{nr}$, make $s$ the subject of the relation

$ A.\:\: s = \dfrac{nrp}{nr + m^2} \\[5ex] B.\:\: s = nr + \dfrac{m^2}{mrp} \\[5ex] C.\:\: s = \dfrac{nrp}{mr} + m^2 \\[5ex] D.\:\: s = \dfrac{nrp}{nr} + m^2 \\[5ex] $

We need to isolate $s$

$ p = s + \dfrac{sm^2}{nr} \\[5ex] Swap \\[3ex] s + \dfrac{sm^2}{nr} = p \\[5ex] \dfrac{snr}{nr} + \dfrac{sm^2}{nr} = p \\[5ex] \dfrac{snr + sm^2}{nr} = p \\[5ex] Multiply\:\:both\:\:sides\:\:by\:\:nr \\[3ex] snr + sm^2 = nrp \\[3ex] Factor\:\:the\:\:LHS\:\:by\:\:GCF \\[3ex] GCF = s \\[3ex] s(nr + m^2) = nrp \\[3ex] Divide\:\:both\:\:sides\:\:by\:\:(nr + m^2) \\[3ex] s = \dfrac{nrp}{nr + m^2} $
(65.) JAMB Make $R$ the subject of the formula $S = \sqrt{\dfrac{2R + T}{2RT}}$

$ A.\:\: R = \dfrac{T}{TS^2 - 1} \\[5ex] B.\:\: R = \dfrac{T}{2(TS^2 - 1)} \\[5ex] C.\:\: R = \dfrac{T}{TS^2 + 1} \\[5ex] D.\:\: R = \dfrac{T}{2(TS^2 + 1)} \\[5ex] $

We need to isolate $R$

$ S = \sqrt{\dfrac{2R + T}{2RT}} \\[5ex] Swap \\[3ex] \sqrt{\dfrac{2R + T}{2RT}} = S \\[5ex] Square\:\:both\:\:sides \\[3ex] \dfrac{2R + T}{2RT} = S^2 \\[5ex] Multiply\:\:both\:\:sides\:\:by\:\:2RT \\[3ex] 2R + T = 2RTS^2 \\[3ex] Swap \\[3ex] 2RTS^2 = 2R + T \\[3ex] Collect\:\:like\:\:terms\:\:in\:\:R \\[3ex] 2RTS^2 - 2R = T \\[3ex] Factor\:\:the\:\:LHS\:\:by\:\:GCF \\[3ex] GCF = 2R \\[3ex] 2R(TS^2 - 1) = T \\[3ex] Divide\:\:both\:\:sides\:\:by\:\:2(TS^2 - 1) \\[3ex] R = \dfrac{T}{2(TS^2 - 1)} $
(66.) JAMB If $gt^2 - k - w = 0$, make $g$ the subject of the formula

$ A.\:\: \dfrac{k + w}{t^2} \\[5ex] B.\:\: \dfrac{k - w}{t^2} \\[5ex] C.\:\: \dfrac{k + w}{t} \\[5ex] D.\:\: \dfrac{k - w}{t} \\[5ex] $

We need to isolate $g$

$ gt^2 - k - w = 0 \\[3ex] Add\:\:k\:\:and\:\:w\:\:to\:\:both\:\:sides \\[3ex] gt^2 = 0 + k + w \\[3ex] gt^2 = k + w \\[3ex] Divide\:\:both\:\:sides\:\:by\:\:t^2 \\[3ex] g = \dfrac{k + w}{t^2} $
(67.) ACT Given that $x = \dfrac{4a + b}{3}$, which of the following expressions is equivalent to $b$

$ F.\:\: 3x - 4a \\[3ex] G.\:\: 3x + 4a \\[3ex] H.\:\: x - \dfrac{4a}{3} \\[5ex] J.\:\: \dfrac{x}{3} - 4a \\[5ex] K.\:\: \dfrac{x - 4a}{3} \\[5ex] $

We need to isolate $b$

$ x = \dfrac{4a + b}{3} \\[5ex] Swap \\[3ex] \dfrac{4a + b}{3} = x \\[5ex] Multiply\:\:both\:\:sides\:\:by\:\:3 \\[3ex] 4a + b = 3x \\[3ex] Subtract\:\:4a\:\:from\:\:both\:\:sides \\[3ex] b = 3x - 4a $
(68.) ACT If $a = 2c$ and $b = 6c$, which of the following relationships holds between $a$ and $b$ for each nonzero value of $c$?

$ A.\:\: a = 3b \\[3ex] B.\:\: a = 2b \\[3ex] C.\:\: a = b \\[3ex] D.\:\: a = \dfrac{1}{6}b \\[5ex] E.\:\: a = \dfrac{1}{3}b \\[5ex] $

We need to express $a$ in terms of $b$

$ a = 2c \\[3ex] b = 6c \\[3ex] 6c = 3 * 2c \\[3ex] \rightarrow b = 3 * a \\[3ex] b = 3a \\[3ex] 3a = b \\[3ex] Divide\:\:both\:\:sides\:\:by\:\:3 \\[3ex] a = \dfrac{b}{3} = \dfrac{1}{3}b $
(69.) ACT Whenever $4x + 7 = 2x - g$, which of the following expressions must be equal to $x$?

$ A.\:\: \dfrac{-g - 7}{2} \\[5ex] B.\:\: \dfrac{-g + 7}{2} \\[5ex] C.\:\: \dfrac{-g + 7}{6} \\[5ex] D.\:\: \dfrac{g}{2} \\[5ex] E.\:\: -\dfrac{7}{4} \\[5ex] $

We need to isolate $x$

$ 4x + 7 = 2x - g \\[3ex] Collect\:\:like\:\:terms\:\:in\:\:x \\[3ex] 4x - 2x = -g - 7 \\[3ex] 2x = -g - 7 \\[3ex] Divide\:\:both\:\:sides\:\:by\:\:2 \\[3ex] x = \dfrac{-g - 7}{2} $
(70.) ACT When solved for $y$, what is $3x + 5y - 13 = 0$?

$ A.\:\: y = \dfrac{-5x + 13}{3} \\[5ex] B.\:\: y = \dfrac{-3x + 13}{5} \\[5ex] C.\:\: y = \dfrac{3x + 13}{5} \\[5ex] D.\:\: y = -3x - 8 \\[3ex] E.\:\: y = -15x - 65 \\[3ex] $

We need to isolate $y$

$ 3x + 5y - 13 = 0 \\[3ex] Subtract\:\:3x\:\:from\:\:both\:\:sides \\[3ex] 5y - 13 = 0 - 3x \\[3ex] 5y - 13 = - 3x \\[3ex] Add\:\:13\:\:to\:\:both\:\:sides \\[3ex] 5y = -3x + 13 \\[3ex] Divide\:\:both\:\:sides\:\:by\:\:5 \\[3ex] y = \dfrac{-3x + 13}{5} $