Systems of Equations Word Problems

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Question 1 of 10

Sue went shopping on two occasions. On the first occasion she bought

5

$5$ apples and

2

$2$ bananas for $2.80. On the second occasion she paid $5.10 for

3

$3$ apples and

5

$5$ bananas. What is the cost of each fruit?

Apple $= $$ $=$$ (0.20)

Banana $= $$ $=$$ (0.90)

Question 2 of 10

A customer bought

4

$4$ drinks and

3

$3$ pizzas for

$ 40.80

$$40.80$ . Another costumer bought

2

$2$ drinks and

1

$1$ pizza for

$ 15.00

$$15.00$ . Find the price of the following:

$(i)$ $(i)$ A drink: $$$ $$$ (2.10)

$(i i)$ $(ii)$ A pizza: $$$ $$$ (10.80)

Question 3 of 10

2

$2$ oranges and

5

$5$ apples cost

$ 5.75

$$5.75$ and

4

$4$ oranges and

6

$6$ apples cost

$ 8.10

$$8.10$ . Find the price of each apple and orange

Orange $=$ $=$ (75) $cents$ $\text(cents)$

Apple $=$ $=$ (85) $cents$ $\text(cents)$

Question 4 of 10

4. Question
At a theatre, there are $400$ $400$ patrons. If there were $60$ $60$ more women than men, how many men attended the show?
- (170)
Hint

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First, let variables to represent men and women.

$x =$ $x=$ number of women

$y =$ $y=$ number of men

Next, write the simultaneous equations being represented in the problem.

$x$ $x$ $+$ $+$ $y$ $y$ $=$ $=$ $400$ $400$ There are a total of $400$ $400$ patrons

$x$ $x$ $=$ $=$ $y$ $y$ $+ 60$ $+60$ $60$ $60$ more women than men

Rewrite equation $2$ $2$ such that $x$ $x$ and $y$ $y$ are on the same side.

$x$ $x$ $=$ $=$ $y$ $y$ $+ 60$ $+60$

$x$ $x$ $- y$ $-y$ $=$ $=$ $y$ $y$ $+ 60$ $+60$ $- y$ $-y$ Subtract $y$ $y$ from both sides

$x$ $x$ $-$ $-$ $y$ $y$ $=$ $=$ $60$ $60$

Add the two equations.

$x$ $x$ $+ 4$ $+4$ $y$ $y$ $=$ $=$ $400$ $400$

$x$ $x$ $-$ $-$ $y$ $y$ $=$ $=$ $60$ $60$

$2 x$ $2x$ $=$ $=$ $460$ $460$

$x$ $x$ $=$ $=$ $230$ $230$ Divide both sides by $2$ $2$

Solve for $y$ $y$ , the number of men.

$x$ $x$ $-$ $-$ $y$ $y$ $=$ $=$ $60$ $60$

$230$ $230$ $-$ $-$ $y$ $y$ $=$ $=$ $60$ $60$ Substitute $x = 230$ $x=230$

$230 - y$ $230-y$ $- 230$ $-230$ $=$ $=$ $60$ $60$ $- 230$ $-230$

$- y$ $-y$ $=$ $=$ $- 170$ $-170$

$y$ $y$ $=$ $=$ $170$ $170$

There were $170$ $170$ men who attended the show.

Question 5 of 10

Find the value of

x

$x$ and

y

$y$

$x =$ $x=$ (5)

$y =$ $y=$ (3)

Question 6 of 10

What are the values of the pronumerals

x

$x$ and

y

$y$ ?

$x =$ $x=$ (45) $c m$ $cm$

$y =$ $y=$ (35) $c m$ $cm$

Question 7 of 10

Find the value of the base angles

$Base Angle =$ $\text(Base Angle) =$ (70) $°$ $°$

Question 8 of 10

8. Question
A stack of $37$ $37$ coins consisting of $10$ $10$ cents and $50$ $50$ cents adds up to $$ 9.70$ $$9.70$ . How many $10$ $10$ cent and $50$ $50$ cent coins are there?
- $10 cents =$ $10 \text(cents)=$ (22)
  
  $50 cents =$ $50 \text(cents)=$ (15)
Hint

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Correct

Well Done!

Incorrect

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First, let $x$ $x$ be the number of $10$ $10$ cents coins and $y$ $y$ be the number of $50$ $50$ cents coins

Use these values to create simultaneous equations

$50$ $50$ $x$ $x$ $+ 50$ $+50$ $y$ $y$ $=$ $=$ $970$ $970$ Total value of coins

$x$ $x$ $+$ $+$ $y$ $y$ $=$ $=$ $37$ $37$ Number of coins

Multiply equation $2$ $2$ by $10$ $10$

$($ $($ $x$ $x$ $+$ $+$ $y$ $y$ $)$ $)$ $\times 10$ $xx10$ $=$ $=$ $37$ $37$ $\times 10$ $xx10$

$10$ $10$ $x$ $x$ $+ 10$ $+10$ $y$ $y$ $=$ $=$ $370$ $370$

Subtract the transformed equation from the first equation.

$10$ $10$ $x$ $x$ $+ 50$ $+50$ $y$ $y$ $=$ $=$ $970$ $970$

$10$ $10$ $x$ $x$ $+ 10$ $+10$ $y$ $y$ $=$ $=$ $370$ $370$

$40 y$ $40y$ $=$ $=$ $600$ $600$

$y$ $y$ $=$ $=$ $15$ $15$ Divide both sides by $40$ $40$

Solve for $x$ $x$ .

$x$ $x$ $+$ $+$ $y$ $y$ $=$ $=$ $37$ $37$

$x$ $x$ $+$ $+$ $15$ $15$ $=$ $=$ $37$ $37$ Substitute $y = 15$ $y=15$

$x + 15$ $x+15$ $- 15$ $-15$ $=$ $=$ $37$ $37$ $- 15$ $-15$ Subtract $15$ $15$ from both sides

$x$ $x$ $=$ $=$ $22$ $22$

$x = 22$ $x =22$ , $y = 15$ $y=15$
Question 9 of 10

9. Question
Jason is three times as old as his son Peter. The sum of their ages is $64$ $64$ . What are their ages?
- Peter's age $=$ $=$ (16)
  
  Jason's age $=$ $=$ (48)
Hint

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Fantastic!

Incorrect

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First, let variables to represent the age of Peter and Jason.

$x =$ $x=$ Peter’s age

$y =$ $y=$ Jason’s age

Next, write the simultaneous equations being represented in the problem.

$y$ $y$ $=$ $=$ $3$ $3$ $x$ $x$ Jason is three times as old as Peter.

$x$ $x$ $+$ $+$ $y$ $y$ $=$ $=$ $64$ $64$ The sum of their ages is $64$ $64$

Substitute Equation $1$ $1$ into $2$ $2$ and solve for Peter’s age $(x)$ $(x)$ .

$x$ $x$ $+$ $+$ $y$ $y$ $=$ $=$ $64$ $64$

$x$ $x$ $+$ $+$ $3 x$ $3x$ $=$ $=$ $64$ $64$ $y = 3 x$ $y=3x$

$4 x$ $4x$ $=$ $=$ $64$ $64$

$x$ $x$ $=$ $=$ $16$ $16$ Divide both sides by $4$ $4$

Solve for Jason’s age $(y)$ $(y)$ .

$y$ $y$ $=$ $=$ $3$ $3$ $x$ $x$

$y$ $y$ $=$ $=$ $3$ $3$ $(16)$ $(16)$ Substitute $x = 16$ $x=16$

$y$ $y$ $=$ $=$ $48$ $48$ Simplify

Peter’s age is $16$ $16$ while Jason is $48$ $48$ years old.
Question 10 of 10

10. Question
Find the value of $x$ $x$ and $y$ $y$
- $x =$ $x=$ (20) $°$ $°$
  
  $y =$ $y=$ (30) $°$ $°$
Hint

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Correct

Great Work!

Incorrect

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First, recall that supplementary angles are equal to $180 °$ $180°$

Equate the values of the two lower angles to $180 °$ $180°$ and simplify

$3 x + 2 y + 3 x$ $3x+2y+3x$ $=$ $=$ $180$ $180$

$6 x + 2 y$ $6x+2y$ $=$ $=$ $180$ $180$ Combine like terms

Next, recall that alternate angles are equal

Equate the alternate angles and simplify

$2 y$ $2y$ $=$ $=$ $3 x$ $3x$

Next, write the simultaneous equations being represented in the problem.

$6$ $6$ $x$ $x$ $+ 2$ $+2$ $y$ $y$ $=$ $=$ $180$ $180$ First equation

$2$ $2$ $y$ $y$ $=$ $=$ $3$ $3$ $x$ $x$ Second equation

Substitute the value of $2 y$ $2y$ into Equation $1$ $1$ and solve for $x$ $x$ .

$6$ $6$ $x$ $x$ $+ 2$ $+2$ $y$ $y$ $=$ $=$ $180$ $180$

$6$ $6$ $x$ $x$ $+$ $+$ $3 x$ $3x$ $=$ $=$ $180$ $180$ $2 y = 3 x$ $2y=3x$

$9 x$ $9x$ $=$ $=$ $180$ $180$

$x$ $x$ $=$ $=$ $20 °$ $20°$ Divide both sides by $9$ $9$

Solve for $y$ $y$ .

$2$ $2$ $y$ $y$ $=$ $=$ $3$ $3$ $x$ $x$

$2$ $2$ $y$ $y$ $=$ $=$ $3$ $3$ $(20)$ $(20)$ Substitute $x = 20$ $x=20$

$2 y$ $2y$ $\div 2$ $divide2$ $=$ $=$ $60$ $60$ $\div 2$ $divide2$ Divide both sides by $2$ $2$

$y$ $y$ $=$ $=$ $30 °$ $30°$

$x = 20 °$ $x =20°$ , $y = 30 °$ $y=30°$

$5$ $5$ $a$ $a$ $+ 2$ $+2$ $b$ $b$	$=$ $=$	$2.80$ $2.80$	First occasion
$3$ $3$ $a$ $a$ $+ 5$ $+5$ $b$ $b$	$=$ $=$	$5.10$ $5.10$	Second occasion

$(5$ $(5$ $a$ $a$ $+ 2$ $+2$ $b$ $b$ $)$ $)$ $\times 3$ $xx3$	$=$ $=$	$2.80$ $2.80$ $\times 3$ $xx3$
$15$ $15$ $a$ $a$ $+ 6$ $+6$ $b$ $b$	$=$ $=$	$8.40$ $8.40$

$(3$ $(3$ $a$ $a$ $+ 5$ $+5$ $b$ $b$ $)$ $)$ $\times 5$ $xx5$	$=$ $=$	$5.10$ $5.10$ $\times 5$ $xx5$
$15$ $15$ $a$ $a$ $+ 25$ $+25$ $b$ $b$	$=$ $=$	$25.50$ $25.50$

$15$ $15$ $a$ $a$ $+ 6$ $+6$ $b$ $b$	$=$ $=$	$8.40$ $8.40$
$15$ $15$ $a$ $a$ $+ 25$ $+25$ $b$ $b$	$=$ $=$	$25.50$ $25.50$

$- 19 b$ $-19b$	$=$ $=$	$- 17.10$ $-17.10$
$b$ $b$	$=$ $=$	$0.90$ $0.90$	Divide both sides by $- 19$ $-19$

$3$ $3$ $a$ $a$ $+ 5$ $+5$ $b$ $b$	$=$ $=$	$5.10$ $5.10$
$3$ $3$ $a$ $a$ $+ 5$ $+5$ $(0.90)$ $(0.90)$	$=$ $=$	$5.10$ $5.10$	Substitute $b = 0.90$ $b=0.90$
$3 a + 4.5$ $3a+4.5$ $- 4.5$ $-4.5$	$=$ $=$	$5.10$ $5.10$ $- 4.5$ $-4.5$	Subtract $4.5$ $4.5$ from both sides
$3 a$ $3a$	$=$ $=$	$0.60$ $0.60$	Divide both sides by $3$ $3$
$a$ $a$	$=$ $=$	$0.20$ $0.20$

Systems of Equations Word Problems

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Quiz summary

Information

1. Question

Hint

Great Work!

2. Question

Hint

Correct!

3. Question

Hint

Correct!

4. Question

Hint

Keep Going!

5. Question

Hint

Fantastic!

6. Question

Hint

Excellent!

7. Question

Hint

Nice Job!

8. Question

Hint

Well Done!

9. Question

Hint

Fantastic!

10. Question

Hint

Great Work!

Quizzes

$4$ $4$ $d$ $d$ $+ 3$ $+3$ $p$ $p$	$=$ $=$	$4080$ $4080$	Equation $1$ $1$
$2$ $2$ $d$ $d$ $+ 1$ $+1$ $p$ $p$	$=$ $=$	$1500$ $1500$	Equation $2$ $2$

$2 d + 1 p$ $2d+1p$	$=$ $=$	$1500$ $1500$	Equation $2$ $2$
$(2 d + 1 p)$ $(2d+1p)$ $\times 2$ $times2$	$=$ $=$	$1500$ $1500$ $\times 2$ $times2$	Multiply the values of both sides by $2$ $2$
$4 d + 2 p$ $4d+2p$	$=$ $=$	$3000$ $3000$	Equation $3$ $3$

$4 d - 3 p$ $4d-3p$	$=$ $=$	$4080$ $4080$
$-$ $-$ $(4 d + 2 p)$ $(4d+2p)$	$=$ $=$	$3000$ $3000$

$p$ $p$	$=$ $=$	$1080$ $1080$	$4 d - 4 d$ $4d-4d$ cancels out

$2 d + 1$ $2d+1$ $p$ $p$	$=$ $=$	$1500$ $1500$	Equation $2$ $2$
$2 d + 1$ $2d+1$ $(1080)$ $(1080)$	$=$ $=$	$1500$ $1500$	$p = 1080$ $p=1080$
$2 d + 1080$ $2d+1080$ $- 1080$ $-1080$	$=$ $=$	$1500$ $1500$ $- 1080$ $-1080$	Subtract $1080$ $1080$ from both sides
$2 d$ $2d$ $\div 2$ $div2$	$=$ $=$	$420$ $420$ $\div 2$ $div2$	Divide both sides by $2$ $2$
$d$ $d$	$=$ $=$	$210$ $210$

$d$ $d$	$=$ $=$	$210 \div 100$ $210div100$
	$=$ $=$	$$ 2.10$ $$2.10$

$p$ $p$	$=$ $=$	$1080 \div 100$ $1080div100$
	$=$ $=$	$$ 10.80$ $$10.80$

$2$ $2$ $x$ $x$ $+ 5$ $+5$ $y$ $y$	$=$ $=$	$575$ $575$	First occasion
$4$ $4$ $x$ $x$ $+ 6$ $+6$ $y$ $y$	$=$ $=$	$810$ $810$	Second occasion

$(2$ $(2$ $x$ $x$ $+ 5$ $+5$ $y$ $y$ $)$ $)$ $\times 2$ $xx2$	$=$ $=$	$575$ $575$ $\times 2$ $xx2$
$4$ $4$ $x$ $x$ $+ 10$ $+10$ $y$ $y$	$=$ $=$	$1150$ $1150$

$x$ $x$ $+$ $+$ $y$ $y$	$=$ $=$	$400$ $400$	There are a total of $400$ $400$ patrons
$x$ $x$	$=$ $=$	$y$ $y$ $+ 60$ $+60$	$60$ $60$ more women than men

$x$ $x$ $+ 4$ $+4$ $y$ $y$	$=$ $=$	$400$ $400$
$x$ $x$ $-$ $-$ $y$ $y$	$=$ $=$	$60$ $60$

$2 x$ $2x$	$=$ $=$	$460$ $460$
$x$ $x$	$=$ $=$	$230$ $230$	Divide both sides by $2$ $2$

$3 x - 17 + x + y + 40$ $3x-17+x+y+40$	$=$ $=$	$180$ $180$
$4 x + y + 23$ $4x+y+23$	$=$ $=$	$180$ $180$	Combine like terms
$4 x + y + 23$ $4x+y+23$ $- 23$ $-23$	$=$ $=$	$180$ $180$ $- 23$ $-23$	Subtract $23$ $23$ from both sides
$4 x + y$ $4x+y$	$=$ $=$	$157$ $157$	Subtract $y$ $y$ from both sides

$3$ $3$ $x$ $x$ $-$ $-$ $y$ $y$	$=$ $=$	$12$ $12$	Length
$2$ $2$ $x$ $x$ $+ 3$ $+3$ $y$ $y$	$=$ $=$	$19$ $19$	Width

$(3$ $(3$ $x$ $x$ $-$ $-$ $y$ $y$ $)$ $)$ $\times 3$ $xx3$	$=$ $=$	$12$ $12$ $\times 3$ $xx3$
$9$ $9$ $x$ $x$ $- 3$ $-3$ $y$ $y$	$=$ $=$	$36$ $36$

$5$ $5$ $x$ $x$ $+ 2$ $+2$ $y$ $y$	$=$ $=$	$295$ $295$	Length
$3$ $3$ $x$ $x$ $+ 4$ $+4$ $y$ $y$	$=$ $=$	$275$ $275$	Width

$(5$ $(5$ $x$ $x$ $+ 2$ $+2$ $y$ $y$ $)$ $)$ $\times 2$ $xx2$	$=$ $=$	$295$ $295$ $\times 2$ $xx2$
$10$ $10$ $x$ $x$ $+ 4$ $+4$ $y$ $y$	$=$ $=$	$590$ $590$

$3 x - 17$ $3x-17$	$=$ $=$	$x + y$ $x+y$
$3 x - 17$ $3x-17$ $- x$ $-x$	$=$ $=$	$x + y$ $x+y$ $- x$ $-x$	Subtract $x$ $x$ from both sides
$2 x - 17$ $2x-17$ $+ 17$ $+17$	$=$ $=$	$y$ $y$ $+ 17$ $+17$	Add $17$ $17$ to both sides
$2 x$ $2x$ $- y$ $-y$	$=$ $=$	$y + 17$ $y+17$ $- y$ $-y$	Subtract $y$ $y$ from both sides
$2 x - y$ $2x-y$	$=$ $=$	$17$ $17$

$2$ $2$ $x$ $x$ $-$ $-$ $y$ $y$	$=$ $=$	$17$ $17$	First equation
$4$ $4$ $x$ $x$ $+$ $+$ $y$ $y$	$=$ $=$	$157$ $157$	Second equation

$x$ $x$ $+$ $+$ $y$ $y$	$=$ $=$	$29$ $29$ $+$ $+$ $41$ $41$	Substitute known values
	$=$ $=$	$70 °$ $70°$

$3$ $3$ $x$ $x$ $- 17$ $-17$	$=$ $=$	$3$ $3$ $(29)$ $(29)$ $- 17$ $-17$	Substitute known values
	$=$ $=$	$87 - 17$ $87-17$
	$=$ $=$	$70 °$ $70°$

$50$ $50$ $x$ $x$ $+ 50$ $+50$ $y$ $y$	$=$ $=$	$970$ $970$	Total value of coins
$x$ $x$ $+$ $+$ $y$ $y$	$=$ $=$	$37$ $37$	Number of coins

$($ $($ $x$ $x$ $+$ $+$ $y$ $y$ $)$ $)$ $\times 10$ $xx10$	$=$ $=$	$37$ $37$ $\times 10$ $xx10$
$10$ $10$ $x$ $x$ $+ 10$ $+10$ $y$ $y$	$=$ $=$	$370$ $370$

$10$ $10$ $x$ $x$ $+ 50$ $+50$ $y$ $y$	$=$ $=$	$970$ $970$
$10$ $10$ $x$ $x$ $+ 10$ $+10$ $y$ $y$	$=$ $=$	$370$ $370$

$40 y$ $40y$	$=$ $=$	$600$ $600$
$y$ $y$	$=$ $=$	$15$ $15$	Divide both sides by $40$ $40$

$y$ $y$	$=$ $=$	$3$ $3$ $x$ $x$	Jason is three times as old as Peter.
$x$ $x$ $+$ $+$ $y$ $y$	$=$ $=$	$64$ $64$	The sum of their ages is $64$ $64$

$3 x + 2 y + 3 x$ $3x+2y+3x$	$=$ $=$	$180$ $180$
$6 x + 2 y$ $6x+2y$	$=$ $=$	$180$ $180$	Combine like terms

$6$ $6$ $x$ $x$ $+ 2$ $+2$ $y$ $y$	$=$ $=$	$180$ $180$	First equation
$2$ $2$ $y$ $y$	$=$ $=$	$3$ $3$ $x$ $x$	Second equation