SSAT Upper Level

Systems of Equations

Q: How do I know whether to use substitution or elimination?

If one equation already has a letter all by itself (like `y = 2x + 1`), substitution is usually the fastest. If both equations have the letters mixed together (like `3x + 2y = 10`), try stacking them for elimination!

Solving for two or more unknowns simultaneously using substitution, elimination, or graphing

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Imagine you are at a carnival snack stand. You buy 2 giant pretzels and 1 soda for $8. Your best friend buys 1 giant pretzel and 2 sodas for $7. 🥨🥤 How much does a single pretzel cost? How about one soda? You don't know the individual prices, but you have two great clues!

Welcome to Systems of Equations! It sounds like a super complicated robot term, but it's really just a fancy math phrase for solving two puzzles at the exact same time. On the SSAT, you will get to be a math detective. Instead of pretzels and sodas, you usually get clues with the letters $x$ and $y$ . A system of equations happens when you have two unknown numbers and two math sentences (equations) that give you hints about them.

To crack the case, you have two superpower moves. Move number one is called Substitution. This is when you know exactly what one letter equals, so you swap it into the other equation like a secret agent in disguise! 🕵️‍♂️ Move number two is Elimination. This is when you stack the two equations on top of each other and add or subtract them so that one of the letters completely vanishes! Poof! 🪄 Once you find the hidden number for the first letter, you just plug it back in to find the second one. You've got this!

Key Formula

To solve by substitution: If

y = 2 x + 1

and

x + y = 10

, replace

y

in the second equation to get

x + (2 x + 1) = 10

. Solve for

x

, then plug it back in to find

y

Practice Questions

7 practice questions for SSAT Upper Level

Q1 Hard

5 x - 2 y = 17

2 x + 2 y = 11

If (x, y) is the solution to the system of equations above, what is the value of

y

1

\frac{3}{2}

2

\frac{5}{2}

4

Show Solution

To find the value of $y$ , first solve for $x$ by adding the two equations together to eliminate $y$ .

$(5 x - 2 y) + (2 x + 2 y) = 17 + 11$
$7 x = 28$
$x = 4$
Now, substitute $x = 4$ back into the second equation:
$2 (4) + 2 y = 11$
$8 + 2 y = 11$
Subtract 8 from both sides:
$2 y = 3$
Divide by 2:
$y = \frac{3}{2}$

Answer: B

Q2 Hard

The sum of two numbers is 84, and their difference is 16. What is the value of the larger number?

A 34

B 42

C 50

D 68

E 100

Show Solution

Let $x$ be the larger number and $y$ be the smaller number. We can translate the problem into a system of equations:

$x + y = 84$
$x - y = 16$
Add the two equations together to eliminate $y$ :
$2 x = 100$
Divide by 2:
$x = 50$
The larger number is 50. (You can verify this by finding $y = 34$ ; their sum is 84 and their difference is 16).

Answer: C

Q3 Hard

2 x + y = 12

and

x - 3 y = - 1

, what is the value of

y

A -2

B 2

C 5

D 7

E 10

Show Solution

We can solve this system using substitution or elimination. Using elimination, multiply the first equation by 3 to easily cancel out the $y$ terms:

$3 (2 x + y) = 3 (12) \Rightarrow 6 x + 3 y = 36$
Add this new equation to the second equation:
$(6 x + 3 y) + (x - 3 y) = 36 + (- 1)$
$7 x = 35$
$x = 5$
Now substitute $x = 5$ into the original second equation:
$5 - 3 y = - 1$
Subtract 5 from both sides:
$- 3 y = - 6$
Divide by -3:
$y = 2$

Answer: B

Q4 Hard

At a local bakery, the total cost of 3 muffins and 2 coffees is $14. If the total cost of 1 muffin and 2 coffees is $10, what is the price of 1 muffin?

A $2

B $3

C $4

D $5

E $6

Show Solution

Let $m$ represent the price of a muffin and $c$ represent the price of a coffee. Set up a system of linear equations based on the word problem:

$3 m + 2 c = 14$
$m + 2 c = 10$
Subtract the second equation from the first equation to eliminate $c$ :
$(3 m + 2 c) - (m + 2 c) = 14 - 10$
$2 m = 4$
Divide by 2:
$m = 2$
The price of 1 muffin is $2.

Answer: A

Q5 Hard

a = - 3

and

b = 4

, what is the value of

a^{2} - 3 b

A -21

B -3

C 3

D 5

E 21

Show Solution

First, substitute the given values $a = - 3$ and $b = 4$ into the expression: $(- 3)^{2} - 3 (4)$ . According to the order of operations (PEMDAS), evaluate the exponent first: $(- 3)^{2} = 9$ . Next, perform the multiplication: $3 (4) = 12$ . Finally, subtract the values: $9 - 12 = - 3$ .

Answer: B

Q6 Hard

x + y = 12

and

x - y = 4

, what is the value of

x y

A 8

B 16

C 32

D 48

E 64

Show Solution

You can solve this system of equations by adding them together. $(x + y) + (x - y) = 12 + 4$ , which simplifies to $2 x = 16$ . Dividing both sides by 2 gives $x = 8$ . Next, substitute $x = 8$ back into the first equation: $8 + y = 12$ . Subtracting 8 from both sides gives $y = 4$ . Finally, multiply $x$ and $y$ to find the value of $x y$ : $8 \times 4 = 32$ .

Answer: C

Q7 Hard

m = 6

and

n = - 2

, what is the value of

\frac{m + 2 n}{m - n}

\frac{1}{4}

\frac{1}{2}

C 1

D 2

E 4

Show Solution

First, substitute $m = 6$ and $n = - 2$ into the numerator: $6 + 2 (- 2) = 6 - 4 = 2$ . Next, substitute the values into the denominator: $6 - (- 2) = 6 + 2 = 8$ . Now divide the evaluated numerator by the evaluated denominator: $\frac{2}{8}$ . Finally, simplify the fraction by dividing the top and bottom by 2 to get $\frac{1}{4}$ .

Answer: A

Tips & Strategies

Use the answers! If you get stuck on the SSAT, just take the multiple-choice answers and plug them into the equations. The correct answer must make BOTH equations true.
Stack your blocks neatly. Before you add or subtract equations for elimination, make sure your $x$ s, $y$ s, and equal signs are lined up perfectly.

Common Mistakes

Watch out for the half-finished trap! You might do all the hard work to find $x$ , circle it, and move on. But wait... the question might have asked for $y$ ! Always re-read what the question is asking for.
Don't forget to distribute the negative sign! If you subtract an entire equation during elimination, make sure you subtract every single piece of it, not just the first letter.

Frequently Asked Questions

How do I know whether to use substitution or elimination?

If one equation already has a letter all by itself (like $y = 2 x + 1$ ), substitution is usually the fastest. If both equations have the letters mixed together (like $3 x + 2 y = 10$ ), try stacking them for elimination!

Will I have to graph these on the SSAT?

Usually, no! The SSAT focuses on solving these with algebra or through word problems. You won't need to draw a graph yourself, but it's helpful to remember that the solution ( $x$ and $y$ ) is the exact point where the two lines cross on a map.

What if I completely forget how to do the algebra during the test?

Don't panic! Use the magic of SSAT multiple choice. Take the answer choices and plug them in for $x$ and $y$ . If they make both math sentences true, you found the winner without doing any heavy algebra!

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