GMAT Math : Solving equations

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Example Questions

Example Question #11 : Solving Equations

Solve for :

4 (x +8) + 13 = 6 (x + 9) - x

Possible Answers:

$x = 4 \frac{1}{2}$

The equation has no solution.

$x = -4 \frac{1}{2}$

x=9

x = -9

Correct answer:

x = -9

Explanation:

4 (x +8) + 13 = 6 (x + 9) - x

$4 \cdot x +4 \cdot 8 + 13 = 6 \cdot x + 6 \cdot 9 - x$

4 x +32 + 13 = 6x + 54 - x

4 x +45 = 5x + 54

4 x +45 -4x -54 = 5x + 54 -4x -54

-9 = x

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Example Question #12 : Solving Equations

Give all real solutions of the following equation:

$x^{4} - 10x^{2} + 9 = 0$

Possible Answers:

The equation has no solution.

$\left \{ -9, -1, 1, 9\right \}$

$\left \{ 1, 3\right \}$

$\left \{1, 9\right \}$

$\left \{ -3, -1, 1, 3\right \}$

Correct answer:

$\left \{ -3, -1, 1, 3\right \}$

Explanation:

By substituting $u = x^{2}$ and, subsequently, $u^{2} =\left ( x^{2} \right )^{2} = x^{4}$ , this equation be rewritten as a quadratic equation, and solved as such:

$x^{4} - 10x^{2} + 9 = 0$

$u^{2} - 10u + 9 = 0$

We are looking to factor the quadratic expression as (u+?)(u+?) , replacing the two question marks with integers with a product of 9 and a sum of ; these integers are -9,-1 .

(u-9)(u-1) = 0

Substitute back:

$(x^{2}-9)(x^{2}-1) = 0$

These factors can themselves be factored as the difference of squares:

(x+3)(x-3)(x+1)(x-1) = 0

Set each factor to zero and solve:

$x-3 = 0 \Rightarrow x = 3$

$x-1 = 0 \Rightarrow x = 1$

$x+1 = 0 \Rightarrow x = - 1$

$x+3 = 0 \Rightarrow x = - 3$

The solution set is $\left \{ -3, -1, 1, 3\right \}$ .

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Example Question #13 : Solving Equations

Solve for :

$x - 7 \sqrt{x} + 12 = 0$

Possible Answers:

The equation has no solution.

$x = 9 \textrm{ or } x = -9$

$x = \sqrt{3} \textrm{ or } x = 2$

$x = 3 \textrm{ or } x = 4$

$x = 9 \textrm{ or } x = 16$

Correct answer:

$x = 9 \textrm{ or } x = 16$

Explanation:

Substitute $u = \sqrt{x}$ , and, subsequently, $u^{2} = x$ , to rewrite this equation as quadratic, then solve by factoring.

$x - 7 \sqrt{x} + 12 = 0$

$u^{2} - 7 u + 12 = 0$

We can rewrite the quadratic expression as (x + ?) (x + ?) , where the question marks are replaced with integers whose product is 12 and whose sum is ; these integers are $-3\ and -4$ .

(u-3) (u-4)= 0

Set each factor to zero and solve for ; then substitute back and solve for :

u-3 = 0

u=3

$\sqrt{x} =3$

x = 9

u-4 = 0

u=4

$\sqrt{x} =4$

x = 16

The solution set, which can be confirmed by substitution, is $\left \{ 9,16\right \}$ .

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Example Question #14 : Solving Equations

Find all real solutions to the following equation:

$\frac{1}{x^{2}} - \frac{8}{x} +12 = 0$

Possible Answers:

$x = - \frac{1}{2} \textrm{ or } x = - \frac{1}{6}$

The equation has no solution.

$x = -6 \textrm{ or } x = -2$

$x = 2 \textrm{ or } x = 6$

$x = \frac{1}{6} \textrm{ or } x = \frac{1}{2}$

Correct answer:

$x = \frac{1}{6} \textrm{ or } x = \frac{1}{2}$

Explanation:

This can be best solved by substituting $u = \frac{1}{x}$ , and, subsequently, $u ^{2}=\left ( \frac{1}{x} \right ) ^{2} =\frac{1}{x ^{2}}$ , then solving the resulting quadratic equation.

$\frac{1}{x^{2}} - \frac{8}{x} + 12 = 0$

$\frac{1}{x^{2}} - 8 \cdot \frac{1}{x} + 12 = 0$

$u^{2} - 8u + 12 = 0$

Factor the expression on the left by finding two integers whose product is 12 and whose sum is :

(u-6)(u-2) = 0

Set each linear binomial factor to 0, solve separately for , and substitute back:

u-6 = 0

u=6

$\frac{1}{x} = 6$

$x = \frac{1}{6}$

u-2 = 0

u=2

$\frac{1}{x} = 2$

$x = \frac{1}{2}$

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Example Question #15 : Solving Equations

The period of a pendulum - that is, the time it takes for the pendulum to swing once and back - varies directly as the square root of its length.

The pendulum of a giant clock is 18 meters long and has period 8.5 seconds. If the pendulum is lengthened to 21 meters, what will its period be, to the nearest tenth of a second?

Possible Answers:

$9.2 \textrm{ sec}$

$7.9 \textrm{ sec}$

$11.2 \textrm{ sec}$

$10.4 \textrm{ sec}$

$11.6 \textrm{ sec}$

Correct answer:

$9.2 \textrm{ sec}$

Explanation:

The variation equation for this situation is

$\frac{P_{1}}{\sqrt{L_{1}}} = \frac{P_{2}}{\sqrt{L}_{2}}$

Set $P_{1} = 8.5 , L_{1} = 18, {L}_{2}= 21$ , and solve for $P_{2}$ ;

$\frac{8.5}{\sqrt{18}} = \frac{P_{2}}{\sqrt{21}}$

$\frac{8.5}{\sqrt{18}} \cdot{\sqrt{21}} = \frac{P_{2}}{\sqrt{21}} \cdot{\sqrt{21}}$

$P_{2} = \frac{8.5}{\sqrt{18}} \cdot{\sqrt{21}} \approx 9.2 \textrm{ sec}$

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Example Question #16 : Solving Equations

$x = 8^{N}$

Which of these expressions is equal to $\log_{32} x$ ?

Possible Answers:

$N\sqrt[3]{N^{2}}$

$\ln \frac{5}{3}N$

$\frac{3}{5} N$

$\frac{5}{3} N$

$\sqrt[5]{N^{3}}$

Correct answer:

$\frac{3}{5} N$

Explanation:

$2^{5} = 32$

$2 = \sqrt[5]{32}$

$8 = 2^{3} = \left ( \sqrt[5]{32} \right ) ^{3} = 32^{ \frac{3}{5} }$

$x = 8^{N}= \left (32^{ \frac{3}{5} } \right \)^ {N } =32^{ \frac{3}{5} N }$

$\log_{32} x =\log_{32}32^{ \frac{3}{5} N } = \frac{3}{5} N$

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Example Question #17 : Solving Equations

Solve for :

| 4x - 18 | - 18 = 57

Possible Answers:

x= -14.25

x=23.25

$x= -14.25 \textrm{ or }x=23.25$

$x= -23.25 \textrm{ or }x=23.25$

$x= -14.25 \textrm{ or }x=14.25$

Correct answer:

$x= -14.25 \textrm{ or }x=23.25$

Explanation:

First, isolate the absolute value expression on one side:

| 4x - 18 | - 18 = 57

| 4x - 18 | -18+ 18 = 57+18

| 4x - 18 | = 75

Rewrite as a compound sentence:

$4x - 18 = - 75 \textrm{ or } 4x - 18 = 75$

Solve each separately:

4x - 18 = - 75

4x - 18+18 = -75+18

4x= - 57

$4x \div 4= - 57 \div 4$

x= - 14.25

4x - 18 = 75

4x - 18+18 = 75+18

4x= 93

$4x \div 4=93 \div 4$

x = 23.25

The solution set is $\left \{ -14.25, 23.25 \right \}$

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Example Question #18 : Solving Equations

Solve for :

| 4x - 18 | + 18 = 57

Possible Answers:

$x=-5.25 \textrm{ or }x=14.25$

x=-14.25

x=14.25

$x=-14.25\textrm{ or }x=14.25$

$x=-5.25 \textrm{ or }x=5.25$

Correct answer:

$x=-5.25 \textrm{ or }x=14.25$

Explanation:

First, isolate the absolute value expression on one side:

| 4x - 18 | + 18 = 57

| 4x - 18 | + 18 -18 = 57-18

| 4x - 18 | = 39

Rewrite as a compound sentence:

$4x - 18 = - 39 \textrm{ or } 4x - 18 = 39$

Solve each separately:

4x - 18 = - 39

4x - 18+18 = - 39+18

4x= - 21

$4x \div 4= - 21 \div 4$

x= - 5.25

4x - 18 = 39

4x - 18+18 = 39+18

4x= 57

$4x \div 4=57 \div 4$

x = 14.25

The solution set is $\left \{ -5.25, 14.25\right \}$ .

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Example Question #332 : Algebra

Give the solution set of the equation:

$\left | x ^{2} - 2x \right | = 3x -6$

Possible Answers:

$x \in \left \{ 2, 3\right \}$

$x \in \left \{ -3, 2, 3\right \}$

$x \in \left \{ -3, 3\right \}$

The equation has no solution.

$x \in \left \{ -3, 2\right \}$

Correct answer:

$x \in \left \{ 2, 3\right \}$

Explanation:

Write this as a compound equation and solve each separately.

$\left | x ^{2} - 2x \right | = 3x -6$

$x ^{2} - 2x = 3x -6 \textrm{ or }x ^{2} - 2x = -\left ( 3x -6 \right )$

$x ^{2} - 2x = 3x -6$

$x ^{2} - 2x - 3x + 6 = 3x -6 - 3x + 6$

$x ^{2} - 5x + 6 =0$

(x-3) (x-2) =0

$x - 3 = 0 \Rightarrow x = 3$

$x - 2 = 0 \Rightarrow x =2$

$x ^{2} - 2x = -\left ( 3x -6 \right )$

$x ^{2} - 2x = -3x +6$

$x ^{2} - 2x+ 3x - 6= -3x +6 + 3x - 6$

$x ^{2} +x - 6= 0$

(x + 3) (x - 2)= 0

$x + 3 = 0 \Rightarrow x = -3$

$x - 2 = 0 \Rightarrow x =2$

This gives us three possible solutions - $\left \{ -3, 2, 3\right \}$ . We check all three.

x = -3

$\left | x ^{2} - 2x \right | = 3x -6$

$\left | \left ( -3\right ) ^{2} - 2 (-3) \right | = 3 (-3) -6$

$\left | \left9 - (-6) \right | = -9-6$

$\left |15| = -15$

15 = -15

This is a false statement so we can eliminate as a false "solution".

x = 2

$\left | x ^{2} - 2x \right | = 3x -6$

$\left |2^{2} - 2 \cdot 2 \right | = 2 \cdot 3 -6$

$\left | \left 4-4 \right | = 6-6$

|0| = 0

0=0

2 is a solution.

x = 3

$\left | x ^{2} - 2x \right | = 3x -6$

$\left | 3 ^{2} - 2 \cdot 3 \right | = 3 \cdot 3 -6$

$\left | \left 9 - 6\right | = 9-6$

$\left |3 | = 3$

3= 3

3 is a solution.

The solution set is $x \in \left \{ 2, 3\right \}$ .

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Example Question #19 : Equations

Solve for :

$4 ^{5x - 7} = 8 ^{2x -1}$

Possible Answers:

x = 2.75

x = 4.25

x = -1.0625

The equation has no solution

x = 0.6875

Correct answer:

x = 2.75

Explanation:

Since $2 ^{2} = 4$ and $2^{3} = 8$ , replace, and use the exponent rules:

$4 ^{5x - 7} = 8 ^{2x -1}$

$\left ( 2 ^{2} \right ) ^{5x - 7} = \left ( 2^{3} \right ) ^{2x -1}$

$2 ^{2 \left ( 5x - 7 \right )}=2 ^{3 \left ( 2x - 1 \right )}$

Set the exponents equal to each other and solve for :

$2 \left ( 5x - 7 \right )=3 \left ( 2x - 1 \right )$

10x -14=6x - 3

10x-6x -14+14=6x -6x - 3+14

4x = 11

$4x \div 4 = 11 \div 4$

x = 2.75

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GMAT Math : Solving equations

Study concepts, example questions & explanations for GMAT Math

All GMAT Math Resources

Example Questions

Example Question #11 : Solving Equations

Example Question #12 : Solving Equations

Example Question #13 : Solving Equations

Example Question #14 : Solving Equations

Example Question #15 : Solving Equations

Example Question #16 : Solving Equations

Example Question #17 : Solving Equations

Example Question #18 : Solving Equations

Example Question #332 : Algebra

Example Question #19 : Equations

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