Solving Equations - SAT Subject Test in Math II

Card 0 of 16

Question

Solve for :

Answer

To solve for , you need to isolate it to one side of the equation. You can subtract the from the right to the left. Then you can add the 6 from the right to the left:

Next, you can factor out this quadratic equation to solve for . You need to determine which factors of 8 add up to negative 6:

$(x \ \ \ \ \ \)(x \ \ \ \ \ \) = 0$

Finally, you set each binomial equal to 0 and solve for :

$x=2 \ \ \ \ \ \ x=4$

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Question

Give the solution set of the equation $x ^{2} - 4x + 11 = 0$ .

Answer

Use the quadratic formula with :

$x = \frac{-b \pm \sqrt{b^{2}-4ac}}{2a}$

$x = \frac{-(-4) \pm \sqrt{(-4)^{2}-4 (1)(11)}}{2 (1)}$

$x = \frac{4 \pm \sqrt{16-44}}{2 }$

$x = \frac{4 \pm \sqrt{-28}}{2 }$

$x = \frac{4 \pm i \sqrt{28}}{2 }$

$x = \frac{4 \pm i \cdot \sqrt{4} \cdot \sqrt{7}}{2 }$

$x = \frac{4 \pm 2i \sqrt{7}}{2 }$

$x = 2 \pm i \sqrt{7}$

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Question

A large tub has two faucets. The hot water faucet, if turned all the way up, can fill the tub in 15 minutes; the cold water faucet can do the same in 9 minutes. Which of the following responses is closest to the time it takes to fill the tub if both faucets are turned all the way up?

Answer

Work problems can be solved by looking at them as rate problems.

The hot faucet can fill up the tub at a rate of 15 minutes per tub, or $\frac{1}{15}$ tub per minute. The cold faucet, similarly, can fill up the tub at a rate of 9 minutes per tub, or $\frac{1}{9}$ tub per minute.

Suppose the tub fills up in minutes. Then, at the end of this time, the hot faucet has filled up $\frac{1}{15} X$ tub, and the cold faucet has filled up $\frac{1}{9} X$ tub, for a total of one tub. We can set up this equation and solve for :

$\frac{1}{15} X + \frac{1}{9} X = 1$

$45 \cdot \left (\frac{1}{15} X + \frac{1}{9} X \right )= 45 \cdot 1$

$45 \cdot \frac{1}{15} X +45 \cdot \frac{1}{9} X = 45$

$8 X \div 8 = 45 \div 8$

Of the five choices, 6 minutes comes closest.

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Question

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

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

Answer

$x^{2}-6x+9$ 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

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

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

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

Either:

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

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

Or:

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

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Question

Give the solution set of the following rational equation:

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

Answer

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 , 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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Question

Solve the equation for y

Answer

First subtract 27 from both sides of the equation

Add 5z to both sides of the equation

Lastly, divide both sides by 5 to get the y by itself

$y=\frac{3+5z}{5}$

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Question

Solve the equation: $\frac{2}{3x}+4=\frac{1}{2x}$

Answer

To isolate the x-variable, we can multiply both sides by the least common denominator.

The least common denominator is . This will eliminate the fractions.

$6x(\frac{2}{3x}+4)=\frac{1}{2x} \cdot 6x$

Subtract 4 on both sides.

Divide by 24 on both sides.

$\frac{24x}{24}=\frac{-1}{24}$

The answer is: $-\frac{1}{24}$

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Question

Solve the equation: $\frac{1}{3}x-\frac{1}{4}x = \frac{1}{5}$

Answer

Find the least common denominator of both sides of the equation, and multiply it on both sides.

The LCD is 60.

$60(\frac{1}{3}x-\frac{1}{4}x) = \frac{1}{5}\times 60$

Combine like-terms on the left.

Divide by 5 on both sides.

$\frac{5x}{5}=\frac{12}{5}$

The answer is: $\frac{12}{5}$

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Question

Solve the equation:

Answer

Distribute the eight through both terms of the binomial.

Add on both sides.

Add 24 on both sides.

Divide by 9 on both sides.

$\frac{9x}{9}=\frac{40}{9}$

The answer is: $\frac{40}{9}$

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Question

Solve the equation:

Answer

Subtract from both sides.

Add 6 on both sides.

The answer is:

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Question

Solve $\sqrt{x^2 + 8} = 2\sqrt{2x-1}$

Answer

First, we want to get everything inside the square roots, so we distribute the :

$\sqrt{x^2 + 8} = \sqrt{4}\sqrt{2x-1}$

$\sqrt{x^2 + 8} = \sqrt{4(2x-1)}$

$\sqrt{x^2 + 8} = \sqrt{8x-4}$

Now we can clear our the square roots by squaring each side:

$(\sqrt{x^2 + 8})^{2} = (\sqrt{8x-4})^2$

Now we can simplify by moving everything to one side of the equation:

Factoring will give us:

So our answers are:

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Question

Solve $6-\sqrt{x}=11$

Answer

Begin by gathering all the constants to one side of the equation:

$-\sqrt{x}=5$

Now multiply by :

$\sqrt{x}=-5$

And finally, square each side:

This might look all fine and dandy, but let's check our solution by plugging it in to the original equation:

$6-\sqrt{25}=11$

So our solution is invalid, and the problem doesn't have a solution.

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Question

Solve the equation:

Answer

Add two on both sides.

Divide by three on both sides.

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

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

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Question

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

Answer

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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Question

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

Answer

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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Question

Solve the equation:

Answer

Add nine on both sides.

Divide by negative six on both sides.

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

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

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