How to find the solution to a rational equation with LCD

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SAT Math › How to find the solution to a rational equation with LCD

Questions 1 - 7

1

Linesmb1

bm/(_m_2 + 1)

–bm/(_m_2 + 1)

–b/(m + 1)

–b/(_m_2 – 1)

b/(_m_2 + 1)

Explanation

Linesmb5

Linesmb4

2

Give the solution set of the rational equation $\frac{(x- 5)^{2}}{(x- 4) ^{2}} = 1$

$\left { \frac{9}{2} \right }$

$\left { 5\right }$

$\left { 4, 5\right }$

$\left { - \frac{1}{2}, \frac{9}{2} \right }$

$\left { - \frac{9}{2}, \frac{9}{2} \right }$

Explanation

Multiply both sides of the equation by the denominator $(x- 4) ^{2}$ :

$\frac{(x- 5)^{2}}{(x- 4) ^{2}} \cdot (x- 4) ^{2} =1 \cdot (x- 4) ^{2}$

$(x- 5)^{2} = (x- 4) ^{2}$

Rewrite both expression using the binomial square pattern:

$x^{2} - 2 \cdot x \cdot 5+ 5^{2} = x^{2} - 2 \cdot x \cdot 4+ 4^{2}$

$x^{2} -10x + 25 = x^{2} - 8x+16$

This can be rewritten as a linear equation by subtracting $x^{2}$ from both sides:

$x^{2} -10x + 25- x^{2} = x^{2} - 8x+16 - x^{2}$

Solve as a linear equation:

$\frac{-2x}{-2 } =\frac{ -9 }{-2}$

$x= \frac{9}{2}$

3

In the equation below, , , and are non-zero numbers. What is the value of in terms of and ?

$\frac{1}{m^3}-\frac{1}{k^2}=\frac{1}{p}$

$m=(\frac{p+k^2}{2})^3$

$m=\frac{p^2k^3}{p+k^2}$

$m=p^{\frac{1}{2}}-k^{\frac{1}{3}}$

$m=(\frac{p+k^2}{pk^2})^{\frac{1}{3}}$

$m=(\frac{pk^2}{p+k^2})^{\frac{1}{3}}$

Explanation

Pkm_7-21-13

Pkm2_7-21-13

4

Rational_5

–2

–1

0

1

2

Explanation

Rational_2

Rational_3

Rational_4

5

Solve for and using elimination:

and

and

and

and

$x=\frac{5}{2}$ and $y=\frac{5}{3}$

Explanation

When using elimination, you need two factors to cancel out when the two equations are added together. We can get the in the first equation to cancel out with the in the second equation by multiplying everything in the second equation by :

Now our two equations look like this:

The can cancel with the , giving us:

These equations, when summed, give us:

Once we know the value for , we can just plug it into one of our original equations to solve for the value of :

6

Solve:

$\frac{10-x}{40}=-5$

Explanation

$\frac{10-x}{40}=-5$

Multiply by on each side

Subtract on each side

Multiply by on each side

7

Solve for x:

$\frac{x+3}{7}=3$

Explanation

The first step is to cancel out the denominator by multiplying both sides by 7:

$7\cdot \frac{x+3}{7}=3\cdot7$

Subtract 3 from both sides to get by itself:

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