SAT Change the Subject

To solve change-the-subject problems on the SAT, you need to know how to:

• manipulate algebraic expressions
• apply the rules of exponents

If $str=x$, $2x=srp$, and $sr \neq 0$, which one of the following is equal to $p$?

(A) $\ \ \frac{t}{2}$
(B) $\ \ 2t$
(C) $\ \ 2tr$
(D) $\ \ 2st$
(E) $\ \ sr(2t-1)$

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Correct Answer: B

Solution 1:

Here we begin with the first equation and use the second one in a substitution.

$$\begin{array}{l l l l} str &=& x &\quad \text{given}\\ 2str &=& 2x &\quad \text{multiply both sides by}\ 2\\ 2str &=& srp &\quad \text{substitute}\ 2x\ \text{with}\ srp\\ 2t &=& p &\quad \text{divide both sides by}\ sr \neq 0\\ \end{array}$$

Solution 2:

We start with the second equation and solve for $x$:

$$\begin{array}{l l l l} 2x &=& srp &\quad \text{given}\\ x &=& \frac{srp}{2} &\quad \text{divide both sides by}\ 2\\ \end{array}$$

Then we substitute $\frac{srp}{2}$ for $x$ in the first equation:

$$\begin{array}{l l l l} str &=& x &\quad \text{given}\\ str &=& \frac{srp}{2} &\quad \text{substitute}\ x \ \text{with}\ \frac{srp}{2}\\ t &=& \frac{p}{2} &\quad \text{divide both sides by}\ sr \neq 0\\ 2t &=& p &\quad \text{multiply both sides by}\ 2\\ \end{array}$$

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Incorrect Choices:

(A)
Refer to Solution 2. If in the last step you divide the left side by $2$, instead of multiply it by $2$, you will get this wrong answer.

$$\begin{array}{l l l l} t &=& \frac{p}{2} &\\ \frac{t}{2} &=& p &\quad \text{mistake: divided left side by}\ 2\\ \end{array}$$

(C) and (D)
If in the last step of Solution 1 you divide the left side only by $s$ or only by $r$, instead of by $sr$, you will get one of these wrong answers.

$$\begin{array}{lcl} &2str = srp &\quad \text{substitute}\ 2x\ \text{with}\ srp\\ \fbox{2tr}=p\ &\text{or}\ \quad \fbox{2st}=p &\quad \text{mistake: divided left side by}\ \fbox{s or r}\\ \end{array}$$

Likewise, if in Solution 2 you divide the left side only by $s$ or only by $r$, instead of by $sr$, you will get one of these wrong answers.

$$\begin{array}{l cl} &str = \frac{srp}{2} &\quad \text{substitute}\ x \ \text{with}\ str\\ \fbox{tr}=\frac{p}{2}\ &\text{or}\ \quad \fbox{st} = \frac{p}{2} &\quad \text{mistake: divided left side by}\ \fbox{s or r}\\ \end{array}$$

(E)
You will get this wrong answer if in the last step of Solution 1 you subtract $sr$ from both sides, instead of divide both sides by $sr$, like this:

$$\begin{array}{l l l l} 2str &=& srp &\quad \text{substitute}\ 2x\ \text{with}\ srp\\ 2str-sr&=&p &\quad \text{mistake: subtracted}\ sr\ \text{instead of divided by}\ sr\\ \end{array}$$

If $x=\frac{3}{5}(x-10y)$, which of the following is an expression for $x$ in terms of $y$?

(A) $\ \ -15y$

(B) $\ \ -6y$

(C) $\ \ -\frac{1}{15}x$

(D) $\ \ -\frac{1}{15}y$

(E) $\ \ 15y$

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Correct Answer: A

Solution 1:

Tip: Follow order of operations.
We manipulate the given equation until we reach an expression for $x$ in terms of $y$. Here, in the first step, we multiply both sides by $\frac{5}{3}$.

$$\begin{array}{r c l l} x &=& \frac{3}{5}(x-10y) &\quad \text{given}\\ \frac{5x}{3} &=& x-10y &\quad \text{multiply both sides by}\ \frac{5}{3}\\ \frac{2x}{3} &=& -10y &\quad \text{subtract}\ x\ \text{from both sides}\\ x &=& -15y &\quad \text{multiply both sides by}\ \frac{3}{2}\\ \end{array}$$

Solution 2:

Tip: Follow order of operations.
Here, in the first step, we use the distributive property.

$$\begin{array}{r c l l} x &=& \frac{3}{5}(x-10y) &\quad \text{given}\\ x &=& \frac{3x}{5}- 6y &\quad \text{use distributive property}\\ \frac{2x}{5} &=& - 6y &\quad \text{subtract}\ \frac{3x}{5}\ \text{from both sides}\\ x &=& -15y &\quad \text{multiply both sides by}\ \frac{5}{2}\\ \end{array}$$

Solution 3:

Tip: Read the entire question carefully.
Because the equation uses the variables $x$ and $y$, the tendency for most is to solve for $y$.

$$\begin{array}{r c l l} x &=& \frac{3}{5}(x-10y) &\quad \text{given}\\ \frac{5x}{3} &=& x-10y &\quad \text{multiply both sides by}\ \frac{5}{3}\\ \frac{2x}{3} &=& -10y &\quad \text{subtract}\ x\ \text{from both sides}\\ -\frac{x}{15} &=& y &\quad \text{divide both sides by}\ -10\\ \end{array}$$

We have expressed $y$ in terms of $x$ but the prompt asks us to express $x$ in terms of $y$. We are one step away:

$$\begin{array}{r c l l} -\frac{x}{15} &=& y &\quad y\ \text{in terms of}\ x\\ x &=& -15y &\quad \text{multiply both sides by}\ -15\\ \end{array}$$

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Incorrect Choices:

(B)
There is an $x$ on both sides of the equation. If you omit the $x$ on the right hand side and solve this equation $x=\frac{3}{5}(-10y)$ instead of $x=\frac{3}{5}(x-10y)$, then you will get this wrong answer.

(C)
Tip: Read the entire question carefully.
Tip: Eliminate obviously wrong answers.
If we solve for $y$, we will get $y=-\frac{1}{15}x$. We will have expressed $y$ in terms of $x$. But the prompt asks us to find an expression for $x$ in terms of$y$. This is the wrong choice.

(D)
Tip: Read the entire question carefully.
If we solve for $y$, we will get $y=-\frac{1}{15}x$. You will make a mistake if you carelessly change the places of $x$ and $y$, like this: $x=-\frac{1}{15}y$. If you have solved for $y$ , multiply both sides by $-15$ next in oder to express $x$ in terms of $y$. Refer to Solution 3.

(E)
Tip: Select the answer with the correct sign!
We show several ways you can get this wrong answer.

If in Solution 1 you forget the negative sign accompanying the $y$ term, as shown:

$$\begin{array}{r c l l} \frac{2x}{5} &=& \fbox{-}10y &\\ x &=& 15y &\quad \text{don't forget the negative sign}\\ \end{array}$$

If in Solution 2 you forget the negative sign accompanying the $y$ term, like this: $$\begin{array}{r c l l} \frac{2x}{5} &=& \fbox{-} 6y &\\ x &=& 15y &\quad \text{don't forget the negative sign}\\ \end{array}$$

If in Solution 3 you forget the negative sign accompanying the $y$ term, like this:

$$\begin{array}{r c l l} \fbox{-}\frac{x}{15} &=& y &\\ x &=& 15y &\quad \text{don't forget the negative sign}\\ \end{array}$$

Or, if you accidentally pick the answer with the wrong sign, you will get this wrong answer.

If $x$ and $y$ are positive numbers and $x^{3} = y^{6}$, what is $y^{-\frac{1}{3}}$ in terms of $x$?

(A) $\ \ x^{-\frac{1}{9}}$

(B) $\ \ x^{-\frac{1}{6}}$

(C) $\ \ x^{\frac{1}{6}}$

(D) $\ \ x^{\frac{1}{2}}$

(E) $\ \ x$

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Correct Answer: B

We present three solutions, showing nearly every step. If you know the Rules of Exponents well, you will be able to do this problem quickly. If not, your solution will be longer, which could lead to more careless mistakes.

Solution 1:

Tip: Know the Rules of Exponents.
We begin with the given equation:

$$\begin{array}{r c l l} x^{3} &=& y^{6} &\quad \text{given}\\ \sqrt[3]{x^{3}} &=& \sqrt[3]{y^{6}} &\quad \text{take the third root of both sides}\\ x^{\frac{3}{3}} &=& y^{\frac{6}{3}} &\quad \text{apply}\ \sqrt[m]{a^{n}}=a^{\frac{n}{m}}\\ x &=& y^{2} &\quad \text{simplify}\\ \sqrt{x} &=& \sqrt{y^{2}} &\quad \text{square root both sides}\\ x^{\frac{1}{2}} &=& y^{\frac{2}{2}} &\quad \text{apply}\ \sqrt[m]{a^{n}}=a^{\frac{n}{m}}\\ x^{\frac{1}{2}} &=& y &\quad \text{simplify}\\ \end{array}$$

We have an expression for $y$ in terms of $x$, but we are looking for $y^{-\frac{1}{3}}$ in terms of $x$. So, we continue with the result from above.

$$\begin{array}{r c l l} y &=& x^{\frac{1}{2}} &\\ \sqrt[3]{y} &=& \sqrt[3]{x^{\frac{1}{2}}} &\quad \text{take the third root of both sides}\\ y^{\frac{1}{3}} &=& x^{\frac{1}{6}} &\quad \text{apply}\ \sqrt[m]{a^{n}}=a^{\frac{n}{m}}\\ \end{array}$$

Now we have an expression for $y^{\frac{1}{3}}$ and we are still looking for $y^{-\frac{1}{3}}$. We manipulate $y^{-\frac{1}{3}}$, and we substitute $y^{\frac{1}{3}}$ with $x^{\frac{1}{6}}$:

$$\begin{array}{r c l l} y^{-\frac{1}{3}} &=& \frac{1}{y^{\frac{1}{3}}} &\quad \text{apply}\ a^{-n}=\frac{1}{a^{n}}\\ &=&\frac{1}{x^{\frac{1}{6}}} &\quad \text{substitute}\ y^{\frac{1}{3}}\ \text{with}\ x^{\frac{1}{6}}\\ &=&\frac{x^{0}}{x^{\frac{1}{6}}} &\quad \text{apply}\ 1=a^{0}\\ &=&x^{-\frac{1}{6}} &\quad \text{apply}\ \frac{a^{m}}{a^{n}}=a^{m-n}\\ \end{array}$$

Solution 2:

Tip: Know the Rules of Exponents.
If you are comfortable with the rules of exponents, you might be in the habit of raising both sides of an equation to a convenient power. This method is shown here.

$$\begin{array}{r c l l} x^{3} &=& y^{6} &\quad \text{given}\\ (x^{3})^{\frac{1}{3}} &=& (y^{6})^{\frac{1}{3}} &\quad \text{raise both sides to the power of}\ \frac{1}{3}\\ x^{\frac{3}{3}} &=& y^{\frac{6}{3}} &\quad \text{apply}\ (a^{n})^{m}=a^{n \cdot m}\\ x &=& y^{2} &\quad \text{simplify}\\ (x)^{\frac{1}{2}} &=& (y^{2})^{\frac{1}{2}} &\quad \text{raise both sides to the power of}\ \frac{1}{2}\\ x^{\frac{1}{2}} &=& y^{\frac{2}{2}} &\quad \text{apply}\ (a^{n})^{m}=a^{n \cdot m}\\ x^{\frac{1}{2}} &=& y &\quad \text{simplify}\\ (x^{\frac{1}{2}})^{-\frac{1}{3}} &=& (y)^{-\frac{1}{3}} &\quad \text{raise both sides to the power of} -\frac{1}{3}\\ x^{-\frac{1}{6}} &=& y^{-\frac{1}{3}} &\quad \text{simplify}\\ \end{array}$$

Solution 3:

Tip: Know the Rules of Exponents.
Tip: Look for short-cuts.
If you are very familiar with the rules of exponents, then you would be able to use the following method.

Currently we have an expression for $y^{6}$ but we are looking for an expression for $y^{-\frac{1}{3}}$. Focusing just on the exponents, we notice that we can get $-\frac{1}{3}$ if we multiply $6$ by $-\frac{1}{18}.$ So, if we raise $y^{6}$ to the power of $-\frac{1}{18},$ we can apply the rule $(a^{m})^{n}=a^{mn}$ and obtain the desired form of $y$. A demonstration follows:

$$\begin{array}{r c l l} x^{3} &=& y^{6} &\quad \text{given}\\ (x^{3})^{-\frac{1}{18}} &=& (y^{6})^{-\frac{1}{18}} &\quad \text{raise both sides to the power of}\ -\frac{1}{18}\\ x^{-\frac{3}{18}} &=& y^{-\frac{6}{18}} &\quad \text{apply}\ (a^{n})^{m}=a^{n \cdot m}\\ x^{-\frac{1}{6}} &=& y^{-\frac{1}{3}} &\quad \text{simplify the exponents}\\ \end{array}$$

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Incorrect Choices:

(A)
Refer to Solution 1. If in the beginning you accidentally square root the right hand side, instead of cube root it, and you keep going, you will get this wrong answer. The mistake is shown below.

$$\begin{array}{r c l l} x^{3} &=& y^{6} &\quad \text{given}\\ \sqrt[3]{x^{3}} &=& \sqrt{y^{6}} &\quad \text{don't accidentally square root the right side}\\ \end{array}$$

(C)
Tip: Be careful with signs!
If you are solving for $(y)^{\frac{1}{3}}$ instead of $(y)^{-\frac{1}{3}}$, you will get this wrong answer.

(D)
If you are solving for $y$ instead of $(y)^{-\frac{1}{3}}$, you will get this wrong answer.

(E)
Tip: The simplest choice may not be the correct one.

If you thought these examples difficult and you need to review the material, these links will help:

• Change the Subject of a Formula
  $\ldots$ express $x$ in terms of $y$ $\ldots$
• Rules of Exponents
  $a^m \times a^n = a^{ m + n }$, $a^n / a^m = a^ { n - m }\ldots$
• Multi-Step Equations-Basic
• Distributive Property
  $a(b+c)=ab+ac$
• Combining Like Terms

• Know the Rules of Exponents.
• Follow order of operations.
• SAT General Tips