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

$5$ apples and

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

$3$ apples and

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

Apple $=$$ (0.20)

Banana $=$$ (0.90)

Question 2 of 10

A customer bought

$4$ drinks and

$3$ pizzas for

$$40.80$ . Another costumer bought

$2$ drinks and

$1$ pizza for

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

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

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

Question 3 of 10

$2$ oranges and

$5$ apples cost

$$5.75$ and

$4$ oranges and

$6$ apples cost

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

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

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

Question 4 of 10

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

$x=$ number of women

$y=$ number of men

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

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

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

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

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

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

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

Add the two equations.

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

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

$2x$ $=$ $460$

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

Solve for $y$ , the number of men.

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

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

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

$-y$ $=$ $-170$

$y$ $=$ $170$

There were $170$ men who attended the show.
Question 5 of 10

5. Question
Find the value of $x$ and $y$
- $x=$ (5)
  
  $y=$ (3)
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First, write the simultaneous equations being represented in the problem.

$3$ $x$ $-$ $y$ $=$ $12$ Length

$2$ $x$ $+3$ $y$ $=$ $19$ Width

Multiply equation $1$ by $3$

$(3$ $x$ $-$ $y$ $)$ $xx3$ $=$ $12$ $xx3$

$9$ $x$ $-3$ $y$ $=$ $36$

Add the second equation to the transformed equation.

$2$ $x$ $+3$ $y$ $=$ $19$

$9$ $x$ $-3$ $y$ $=$ $36$

$11x$ $=$ $55$

$x$ $=$ $5$ Divide both sides by $11$

Solve for $y$ .

$2$ $x$ $+3$ $y$ $=$ $19$

$2$ $(5)$ $+3$ $y$ $=$ $19$ Substitute $x=5$

$10+3y$ $-10$ $=$ $19$ $-10$ Subtract $10$ from both sides

$3y$ $=$ $9$ Divide both sides by $3$

$y$ $=$ $3$

$x =5$ , $y=3$

Question 6 of 10

What are the values of the pronumerals

$x$ and

$y$ ?

$x=$ (45) $cm$

$y=$ (35) $cm$

Question 7 of 10

Find the value of the base angles

$\text(Base Angle) =$ (70) $°$

Question 8 of 10

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

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

Use these values to create simultaneous equations

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

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

Multiply equation $2$ by $10$

$($ $x$ $+$ $y$ $)$ $xx10$ $=$ $37$ $xx10$

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

Subtract the transformed equation from the first equation.

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

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

$40y$ $=$ $600$

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

Solve for $x$ .

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

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

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

$x$ $=$ $22$

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

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

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

$x=$ Peter’s age

$y=$ Jason’s age

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

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

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

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

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

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

$4x$ $=$ $64$

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

Solve for Jason’s age $(y)$ .

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

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

$y$ $=$ $48$ Simplify

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

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

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

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

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

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

Next, recall that alternate angles are equal

Equate the alternate angles and simplify

$2y$ $=$ $3x$

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

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

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

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

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

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

$9x$ $=$ $180$

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

Solve for $y$ .

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

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

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

$y$ $=$ $30°$

$x =20°$ , $y=30°$

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

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

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

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

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

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

Systems of Equations Word Problems

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Quiz summary

Information

1. Question

Hint

Great Work!

2. Question

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Correct!

3. Question

Hint

Correct!

4. Question

Hint

Keep Going!

5. Question

Hint

Fantastic!

6. Question

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Excellent!

7. Question

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Nice Job!

8. Question

Hint

Well Done!

9. Question

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Fantastic!

10. Question

Hint

Great Work!

Quizzes

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

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

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

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

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

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

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

$2$ $x$ $+5$ $y$	$=$	$575$
$2$ $x$ $+5$ $(85)$	$=$	$575$	Substitute $y=85$
$2x+425$ $-425$	$=$	$575$ $-425$	Subtract $425$ from both sides
$2x$	$=$	$150$	Divide both sides by $2$
$x$	$=$	$75$

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

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

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

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

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

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

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

$x$ $-$ $y$	$=$	$60$
$230$ $-$ $y$	$=$	$60$	Substitute $x=230$
$230-y$ $-230$	$=$	$60$ $-230$
$-y$	$=$	$-170$
$y$	$=$	$170$

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

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

$2$ $x$ $+3$ $y$	$=$	$19$
$2$ $(5)$ $+3$ $y$	$=$	$19$	Substitute $x=5$
$10+3y$ $-10$	$=$	$19$ $-10$	Subtract $10$ from both sides
$3y$	$=$	$9$	Divide both sides by $3$
$y$	$=$	$3$

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

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

$3$ $x$ $+4$ $y$	$=$	$275$
$3$ $(45)$ $+4$ $y$	$=$	$275$	Substitute $x=45$
$135+4y$ $-135$	$=$	$275$ $-135$	Subtract $135$ from both sides
$4y$	$=$	$140$	Divide both sides by $4$
$y$	$=$	$35 cm$

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

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

$2$ $x$ $-$ $y$	$=$	$17$
$2$ $(29)$ $-$ $y$	$=$	$17$	Substitute $x=29$
$58-y$ $-58$	$=$	$17$ $-58$	Subtract $58$ from both sides
$-y$ $times(-1)$	$=$	$-41$ $times(-1)$	Multiply both sides by $-1$
$y$	$=$	$41$

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

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

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

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

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

$x$ $+$ $y$	$=$	$37$
$x$ $+$ $15$	$=$	$37$	Substitute $y=15$
$x+15$ $-15$	$=$	$37$ $-15$	Subtract $15$ from both sides
$x$	$=$	$22$

$x$ $+$ $y$	$=$	$64$
$x$ $+$ $3x$	$=$	$64$	$y=3x$
$4x$	$=$	$64$
$x$	$=$	$16$	Divide both sides by $4$

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

$6$ $x$ $+2$ $y$	$=$	$180$
$6$ $x$ $+$ $3x$	$=$	$180$	$2y=3x$
$9x$	$=$	$180$
$x$	$=$	$20°$	Divide both sides by $9$