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Systems of Equations Word Problems 1Systems of Equations Word Problems 1
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Question 1 of 5
1. Question
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)
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First, let variables to represent apples and bananas.`a=` cost of each apple`b=` cost of each bananaNext, write the systems of equations being represented in the problem.`5``a``+2``b` `=` `2.80` First occasion `3``a``+5``b` `=` `5.10` Second occasion Multiply equation `1` by `3``(5``a``+2``b``)``xx3` `=` `2.80``xx3` `15``a``+6``b` `=` `8.40` Multiply equation `2` by `5``(3``a``+5``b``)``xx5` `=` `5.10``xx5` `15``a``+25``b` `=` `25.50` Subtract the transformed equations.`15``a``+6``b` `=` `8.40` `15``a``+25``b` `=` `25.50` `19b` `=` `17.10` `b` `=` `0.90` Divide both sides by `19` Solve for `a`, the cost of each apple.`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` Apple `=$0.20`, Banana `=$0.90` 

Question 2 of 5
2. Question
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)
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First, let variables represent drinks and pizzas.`d=` cost of each drink`p=` cost of each pizzaNext, write the both of the customer’s purchases as systems of equations.To make the solution easier, convert the dollar value into cents first..`4``d``+3``p` `=` `4080` Equation `1` `2``d``+1``p` `=` `1500` Equation `2` Next, multiply the values of equation `2` by `2` and label the product as equation `3`.`2d+1p` `=` `1500` Equation `2` `(2d+1p)``times2` `=` `1500``times2` Multiply the values of both sides by `2` `4d+2p` `=` `3000` Equation `3` Then, subtract equation `3` from equation `1`.`4d3p` `=` `4080` `` `(4d+2p)` `=` `3000` `p` `=` `1080` `4d4d` cancels out Now, substitute the value of `p` into any of the two equations.`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` Finally, convert the values back to dollars by dividing each them by `100`Dollar value of each drink`d` `=` `210div100` `=` `$2.10` Dollar value of each pizza`p` `=` `1080div100` `=` `$10.80` `\text(Price of each drink)=$2.10``\text(Price of each pizza)=$10.80` 

Question 3 of 5
3. Question
`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)`
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First, let variables represent apples and oranges.`x=` cost of each orange`y=` cost of each appleNext, write the systems of equations being represented in the problem.`2``x``+5``y` `=` `575` First occasion `4``x``+6``y` `=` `810` Second occasion Multiply equation `1` by `2``(2``x``+5``y``)``xx2` `=` `575``xx2` `4``x``+10``y` `=` `1150` Subtract the second equation from the transformed equation.`4``x``+10``y` `=` `1150` `4``x``+6``y` `=` `810` `4y` `=` `340` `y` `=` `85` Divide both sides by `4` Solve for `x`, the cost of each orange.`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` Orange `=75 \text(cents)`, Apple `=85 \text(cents)` 

Question 4 of 5
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 menNext, write the systems of 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` `230y``230` `=` `60``230` `y` `=` `170` `y` `=` `170` There were `170` men who attended the show. 
Question 5 of 5
5. Question
Find the value of `x` and `y`
`x=` (5)`y=` (3)
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First, write the systems of 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` 
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