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SAT II Math II : Solving Equations

Study concepts, example questions & explanations for SAT II Math II

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Example Questions

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Example Question #161 : Sat Subject Test In Math Ii

Solve the equation: 3x-2 = 50

Possible Answers:

$\frac{52}{3}$

Correct answer:

$\frac{52}{3}$

Explanation:

Add two on both sides.

3x-2+2 = 50+2

3x = 52

Divide by three on both sides.

$\frac{3x }{3}=\frac{ 52}{3}$

The answer is: $\frac{52}{3}$

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Example Question #171 : Sat Subject Test In Math Ii

Solve: $\frac{2}{3}x = \frac{5}{2}$

Possible Answers:

$\frac{3}{5}$

$\frac{10}{3}$

$\frac{15}{4}$

$\frac{4}{15}$

$\frac{5}{3}$

Correct answer:

$\frac{15}{4}$

Explanation:

To isolate the x-variable, multiply both sides by the coefficient of the x-variable.

$\frac{2}{3}x \cdot \frac{3}{2} = \frac{5}{2}\cdot\frac{3}{2}$

The answer is: $\frac{15}{4}$

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Example Question #11 : Single Variable Algebra

Give the solution set of the following rational equation:

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

Possible Answers:

$\left \{ 0\right \}$

$\left \{ 0, 4 \right \}$

$\left \{ 0, 2, 4 \right \}$

$\left \{ 2, 4 \right \}$

No solution

Correct answer:

No solution

Explanation:

Multiply both sides of the equation by $x^{2}-2x$ to eliminate the fraction:

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

$\frac{ x^{2}-4x}{x^{2}-2x}\cdot (x^{2}-2x) = 2 \cdot (x^{2}-2x)$

$x^{2}-4x = 2 x^{2}-4x$

Subtract $x^{2}-4x$ from both sides:

$x^{2}-4x- (x^{2}-4x) = 2 x^{2}-4x - (x^{2}-4x)$

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

$x^{2} = 0$

The only possible solution is x = 0 , However, if this is substituted in the original equation, the expression at left is undefined, as seen here:

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

$\frac{ 0^{2}-4 \cdot 0}{0^{2}-2 \cdot 0} = 2$

$\frac{ 0}{ 0} = 2$

An expression with a denominator of 0 has an undefined value, so this statement is false. The equation has no solution.

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Example Question #12 : Solving Equations

Give the set of all real solutions of the following equation:

$\frac{15 }{x^{2}-6x+9} - \frac{11}{x-3}+ 2= 0$

Possible Answers:

None of these

$\left \{ 3 \frac{1}{3}, 3 \frac{2}{5} \right \}$

$\left \{ \frac{1}{6}, \frac{2}{11} \right \}$

$\left \{ 2 \frac{1}{2}\right \}$

$\left \{ 5 \frac{1}{2} , 6 \right \}$

Correct answer:

$\left \{ 5 \frac{1}{2} , 6 \right \}$

Explanation:

$x^{2}-6x+9$ can be seen to fit the perfect square trinomial pattern:

$x^{2}-6x+9 = x^{2} - 2 \cdot x \cdot 3 + 3 ^{2} = (x-3) ^{2}$

The equation can therefore be rewritten as

$\frac{15 }{ (x-3) ^{2}} - \frac{11}{x-3}+ 2= 0$

Multiply both sides of the equation by the least common denominator of the expressions, which is $LCM (x-3 , (x-3) ^{2}) = (x-3)^{2}$ :

$\left [\frac{15 }{ (x-3) ^{2}} - \frac{11}{x-3}+ 2\right ](x-3)^{2}= 0(x-3) ^{2}$

$\frac{15 }{ (x-3) ^{2}} \cdot (x-3)^{2} - \frac{11}{x-3} \cdot (x-3)^{2} + 2\cdot (x-3)^{2} = 0$

$15 -11(x-3 )+ 2\cdot (x^{2} -6x+9 ) = 0$

$15 -11x+33 + 2 x^{2} -12x+18 = 0$

$2 x^{2} -23x +66= 0$

This can be solved using the method. We are looking for two integers whose sum is and whose product is $2 \cdot 66 = 132$ . Through some trial and error, the integers are found to be and , so the above equation can be rewritten, and solved using grouping, as

$2 x^{2} -12x - 11x +66= 0$

$(2 x^{2} -12x) - (11x -66)= 0$

2x (x-6 ) - 11 (x-6 )= 0

(2x - 11 )(x-6 )= 0

By the Zero Product Principle, one of these factors is equal to zero:

Either:

2x-11 = 0

2x-11+ 11 = 0 + 11

2x = 11

$\frac{2x }{2}= \frac{11}{2}$

$x = 5 \frac{1}{2}$

Or:

x - 6 = 0

x - 6 + 6 = 0 + 6

x = 6

Both solutions can be confirmed by substitution; the solution set is $\left \{ 5 \frac{1}{2} , 6 \right \}$ .

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Example Question #172 : Sat Subject Test In Math Ii

Solve: $-3x = \frac{1}{5}$

Possible Answers:

$-\frac{3}{5}$

$-\frac{1}{15}$

$-\frac{5}{3}$

Correct answer:

$-\frac{1}{15}$

Explanation:

To solve for x, multiply by negative one-third on both sides.

$-3x \times -\frac{1}{3} = \frac{1}{5} \times -\frac{1}{3}$

The answer is: $-\frac{1}{15}$

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Example Question #12 : Solving Equations

Solve the equation: -6x-9 = 11

Possible Answers:

$\frac{1}{3}$

$-\frac{20}{3}$

$-\frac{1}{6}$

$-\frac{10}{3}$

$-\frac{1}{3}$

Correct answer:

$-\frac{10}{3}$

Explanation:

Add nine on both sides.

-6x-9 +9= 11+9

-6x = 20

Divide by negative six on both sides.

$\frac{-6x }{-6}=\frac{ 20}{-6}$

The answer is: $-\frac{10}{3}$

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SAT II Math II : Solving Equations

Study concepts, example questions & explanations for SAT II Math II

All SAT II Math II Resources

Example Questions

Example Question #161 : Sat Subject Test In Math Ii

Example Question #171 : Sat Subject Test In Math Ii

Example Question #11 : Single Variable Algebra

Example Question #12 : Solving Equations

Example Question #172 : Sat Subject Test In Math Ii

Example Question #12 : Solving Equations

All SAT II Math II Resources

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