# SAT Algebraic Manipulations

To successfully manipulate algebraic expressions on the SAT, you need to know how to:

- apply addition, subtraction, multiplication, and division to algebraic expressions
- apply the order of operations
- make simple substitutions

## Examples

If \(3m\) + \(6m\) + \(9m\) = \(-36\), which of the following is equal to \(m\)?

(A) \(\ \ 2\)

(B) \(\ \ 1\)

(C) \(\ \ 0\)

(D) \(\ \ -1\)

(E) \(\ \ -2\)

Correct Answer: E

Solution 1:

Tip: Follow order of operations.

Using the given equation, we solve for \(m\).\[\begin{array}{l c l l l} 3m + 6m + 9m &=& -36 &\quad \text{original expression} &(1)\\ 18m &=& -36 &\quad \text{combine like terms} &(2)\\ \frac{18m}{18} &=& \frac{-36}{18} &\quad \text{divide both sides by}\ 18 &(3)\\ m &=& -2 &\quad \text{perform division} &(4)\\ \end{array}\]

Solution 2:

Tip: Plug and check.

We plug the value of each answer choice into the given equation and select the one that doesn't yield a contradiction.(A) If \(m=2\):

\(3m + 6m + 9m = 3 \cdot 2 + 6 \cdot 2 + 9 \cdot 2 = 6 +12 +18 = 36 \neq -36.\)

This is a contradiction. Eliminate (A).

(B) If \(m=1\):

\(3m + 6m + 9m = 3 \cdot 1 + 6 \cdot 1 + 9 \cdot 1 = 3 +6 +9 = 18 \neq -36.\)

This is a contradiction. Eliminate (B).

(C) If \(m=0\):

\(3m + 6m + 9m = 3 \cdot 0 + 6 \cdot 0 + 9 \cdot 0 = 0 \neq -36.\)

This is a contradiction. Eliminate (C).

(D) If \(m=-1\):

\(3m + 6m + 9m = 3 \cdot (-1) + 6 \cdot (-1) + 9 \cdot (-1) = -3 -6 -9 = -18 \neq -36.\)

This is a contradiction. Eliminate (D).

(E) If \(m=-2\):

\(3m + 6m + 9m = 3 \cdot (-2) + 6 \cdot (-2) + 9 \cdot (-2) = -6 -12 -18 = -36.\)

This is correct and therefore (E) is the answer.

Incorrect Choices:

(A)

Tip: Select the answer with the correct sign!

(B),(C), and(D)

Solution 2 explains why these choices are wrong.

If \(-3(2x-5)+8 = -2x+3\), what is the value of \(x\)?

(A) \(\ -5\)

(B) \(\ -\frac{5}{4}\)

(C) \(\ \ 0\)

(D) \(\ \ \frac{5}{4}\)

(E) \(\ \ 5\)

Correct Answser: E

Solution 1:

Tip: Follow order of operations.

We start with the given equation and we simplify.\[\begin{array}{rcll} -3(2x-5)+8 &=& -2x+3 &\text{original equation}&\quad (1)\\ -6x+15+8 &=& -2x+3 &\text{use distributive property}&\quad (2)\\ -6x+23 &=& -2x+3 &\text{simplify}&\quad (3)\\ -6x+23-3&=&-2x+3-3 &\text{subtract} \ 3\ \text{from both sides}&\quad (4)\\ -6x+20 &=&-2x &\text{simplify}&\quad (5)\\ -6x +20+6x &=&-2x +6x&\text{add}\ 6x\ \text{to both sides}&\quad (6)\\ 20 &=& 4x &\text{combine like terms}&\quad (7)\\ \frac{20}{4} &=&\frac{4x}{4}&\text{divide both sides by} \ 4&\quad (8)\\ 5&=&x&\text{simplify the fractions}&\quad (9)\\ \end{array}\]

Solution 2:

Tip: Plug and check.

We can plug each answer choice into the given equation and check if it yields a true statement. If it does, then the choice is right. In this case, only (E) will work.

Incorrect Choices:

(A)

Tip: Select the answer with the correct sign!

Refer to the solution above. The answer should be \(5\). Selecting \(-5\) would be a careless mistake.

(B)

Tip: When distributing, be careful with signs!

Refer to Solution 1 above. If in step \((2)\) we forget to distribute the negative sign, we will get:\[\begin{array}{lcl} \fbox{-}3(2x-5)+8 &=& -2x+3 &\text{original equation}&\quad (1)\\ 6x-15+8 &=& -2x+3 &\text{mistake: didn't distribute}\\ &&&\text{negative sign}&\quad (2)\\ \end{array}\]

Simplifying, we get:

\[\begin{array}{lcl} -\frac{5}{4}&=&x&\text{simplify} \end{array}\] But this is wrong.

(C)

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

Plug in and check. If \(x=0\), we get:\[\begin{array}{lcl} -3(2x-5)+8 &=& -2x+3\\ -3(2\cdot0-5)+8&=&-2\cdot 0+3&\text{plug in}\ x=0\\ -3(0-5)+8&=&=0+3&\text{simplify}\\ -3(-5)+8&=&3&\text{simplify parentheses}\\ 15+8&=&3&\text{simplify the left side}\\ 23&=&3&\text{simplify the left side again}\\ \end{array}\]

But \(23\neq3\). Therefore, this choice is wrong.

(D)

It is possible you made a mistake when reducing a fraction. Refer to step \((8)\) in the solution above and focus on the fraction on the left side of the equation.\[\begin{array}{lcll} \frac{20}{4} &=&\frac{4x}{4}&\text{divide both sides by} \ 4&\quad (8)\\ \end{array}\]

We must divide both the numerator and denominator by their greatest common factor to obtain the correct reduced fraction. \(4\) is the greatest number that divides both \(20\) and \(4\). So, \(\frac{20/4}{4/4}=\frac{5}{1}=5\). But if we forget to divide the denominator by \(4\), we will get this wrong answer.

If \(5x+2=9\), what is the value of \(5x-2\)?

(A) \(\ \ -9\)

(B) \(\ \ -5\)

(C) \(\ \ \frac{7}{5}\)

(D) \(\ \ 5\)

(E) \(\ \ 7\)

Correct Answer: D

Solution 1:

Tip: Look for short-cuts.

We don't need to solve for \(x\) to find the answer. That's the trick. We realize that \(5x-2=5x+2-4=9-4=5\).

Solution 2:We could solve for \(x\):

\[\begin{array}{rcl} 5x+2&=&9&\quad\text{given}\\ 5x&=&7&\quad\text{subtract}\ \ 2\ \text{from both sides}\\ x&=&\frac{7}{5}&\quad\text{divide both sides by}\ 5\\ \end{array}\]

Then, \(5x-2=5\times\frac{7}{5}-2=7-2=5.\)

So, \(5x-2=5\).

Incorrect Choices:

(A)

Tip: If you can, verify your choice.

We are given \(5x+2=9\). We are looking for \(5x-2\), and you may think that because the sign between \(5x\) and \(2\) changed, you need to change the sign of \(9\) also in order to get the answer, like this: \(5x-2=-9\). But verify your choice. If \(5x-2=-9\), then \(5x=-7\) and adding \(2\) to both sides of this equation, we get \(5x+2 = -7+2=-5\neq 9\). Therefore, (C) is the wrong choice.

(B)

Tip: Select the answer with the correct sign!

(C)

Tip: Read the entire question carefully.

You likely got this answer because you solved for \(x\) instead of \(5x-2\).

(E)

Tip: Read the entire question carefully.

You likely got this answer because you solved for \(5x\), not for \(5x-2\).

## Review

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

## SAT Tips for Algebraic Manipulation

- Follow order of operations.
- Select the answer with the correct sign!
- When distributing, be careful with signs!
- SAT General Tips

**Cite as:**SAT Algebraic Manipulations.

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