Applications of Systems of Equations Help

Applications of Systems of Equations

Systems of two linear equations can be used to solve many kinds of word problems. In these problems, two facts will be given about two variables. Each pair of facts can be represented by a linear equation. This gives us a system of two equations with two variables.

Examples

A movie theater charges $4 for each children’s ticket and $6.50 for each adult’s ticket. One night 200 tickets were sold, amounting to $1100 in ticket sales. How many of each type of ticket was sold?

Let x represent the number of children’s tickets sold and y , the number of adult tickets sold. One equation comes from the fact that a total of 200 adult and children’s tickets were sold, giving us x + y = 200. The other equation comes from the fact that the ticket revenue was $1100. The ticket revenue from children’s tickets is 4 x , and the ticket revenue from adult tickets is 6.50y. Their sum is 1100 giving us 4 x + 6.50 y = 1100.

Systems of Equations and Inequalities Examples

We could use either substitution or addition to solve this system. Substitution is a little faster. We will solve for x in B.

x = 200 − y

4(200 − y ) + 6.50 y = 1100 Put 200 − y into A

800 − 4 y + 6.50 y = 1100

y = 120

x = 200 − y = 200 − 120 = 80

Eighty children’s tickets were sold, and 120 adult tickets were sold.

A farmer had a soil test performed. He was told that his field needed 1080 pounds of Mineral A and 920 pounds of Mineral B. Two mixtures of fertilizers provide these minerals. Each bag of Brand I provides 25 pounds of Mineral A and 15 pounds of Mineral B. Brand II provides 20 pounds of Mineral A and 20 pounds of Mineral B. How many bags of each brand should he buy?

Let x represent the number of bags of Brand I and y represent the number of bags of Brand II. Then the number of pounds of Mineral A he will get from Brand I is 25 x and the number of pounds of Mineral B is 15 x . The number of pounds of Mineral A he will get from Brand II is 20 y and the number of pounds of Mineral B is 20 y . He needs 1080 pounds of Mineral A, 25 x pounds will come from Brand I and 20 y will come from Brand II. This gives us the equation 25 x + 20 y = 1080. He needs 920 pounds of Mineral B, 15 x will come from Brand I and 20 y will come from Brand II. This gives us the equation 15 x + 20 y = 920.

Systems of Equations and Inequalities Examples

We will compute A − B.

Systems of Equations and Inequalities Examples

He needs 16 bags of Brand I and 34 bags of Brand II.

A furniture manufacturer has some discontinued fabric and trim in stock. He can use them on sofas and chairs. There are 160 yards of fabric and 110 yards of trim. Each sofa takes 6 yards of fabric and 4.5 yards of trim. Each chair takes 4 yards of fabric and 2 yards of trim. How many sofas and chairs should be produced in order to use all the fabric and trim?

Let x represent the number of sofas to be produced and y , the number of chairs. The manufacturer needs to use 160 yards of fabric, 6 x will be used on sofas and 4 y yards on chairs. This gives us the equation 6 x + 4 y = 160. There are 110 yards of trim, 4.5 x yards will be used on the sofas and 2 y on the chairs. This gives us the equation 4.5 x + 2 y = 110.

Systems of Equations and Inequalities Examples

We will compute F − 2T.

Systems of Equations and Inequalities Examples

The manufacturer needs to produce 20 sofas and 10 chairs.

Applications of Systems of Equations Practice Problems

Practice

For Problems 1-9, solve the systems of equations. Put your solutions in the form of a point, ( x, y ).

A grocery store sells two different brands of milk. The price for the name brand is $3.50 per gallon, and the price for the store’s brand is $2.25 per gallon. On one Saturday, 4500 gallons of milk were sold for sales of $12,875. How many of each brand were sold?
A gardener wants to add 39 pounds of Nutrient A and 16 pounds of Nutrient B to her garden. Each bag of Brand X provides 3 pounds of Nutrient A and 2 pounds of Nutrient B. Each bag of Brand Y provides 4 pounds of Nutrient A and 1 pound of Nutrient B. How many bags of each brand should she buy?
A clothing manufacturer has 70 yards of a certain fabric and 156 buttons in stock. It manufacturers jackets and slacks that use this fabric and button. Each jacket requires yards of fabric and 4 buttons. Each pair of slacks required yards of fabric and 3 buttons. How many jackets and pairs of slacks should the manufacturer produce to use all the available fabric and buttons?

Solutions

Solve for x in B: x = −3 + 2 y and substitute this for x in A.

2 x + 3 y = 1

2(−3 + 2 y ) + 3 y = 1

−6 + 4 y + 3 y = 1

7 y = 7

y = 1 Put y = 1 in x = −3 + 2 y

x = −3 + 2(1) = −1

The solution is (−1, 1).
Solve for x in B: x = −4 y and substitute this for x in A.

x + y = 3

−4 y + y = 3

−3 y = 3

y = −1 Put y = − 1 in x = −4 y

x = −4(−1) =4

The solution is (4, −1).
We will add A + B.

The solution is (1, 3).
The solution is (1, 6).
We will add −A + B.

The solution is (−2, 3).
We will compute A − 2B.

The solution is (1, 1).
We will compute 3A − 5B.

The solution is (2, 4).
We will compute 3A − 2B.

The solution is (1, −3).
First clear the fractions.
Let x represent the number of gallons of the name brand sold and y represent the number of gallons of the store brand sold. The total number of gallons sold is 4500, giving us x + y = 4500. Revenue from the name brand is 3.50x and is 2.25 y for the store brand. Total revenue is $12,875, giving us the equation 3.50 x + 2.25 y = 12,875.

We will use substitution.

x = 4500 − y

3.50(4500 − y ) + 2.25 y = 12,875

y = 2300

x = 4500 − y = 4500 − 2300 = 2200

The store sold 2200 gallons of the name brand and 2300 gallons of the store brand.
Let x represent the number of bags of Brand X and y , the number of bags of Brand Y. She will get 3 x pounds of Nutrient A from x bags of Brand X and 4 y pounds from y bags of Brand Y, so we need 3 x + 4 y = 39. She will get 2 x pounds of Nutrient B from x bags of Brand X and 1 y pounds of Nutrient B from y bags of Brand Y, so we need 2 x + y = 16. We will use substitution.

y = 16 − 2 x

3 x + 4(16 − 2 x ) = 39

x = 5

y = 16 − 2 x = 16 − 2(5) = 6

The gardener needs to buy 5 bags of Brand X and 6 bags of Brand Y.
Let x represent the number of jackets to be produced and y the number of pairs of slacks. To use 70 yards of fabric, we need . To use 156 buttons, we need 4 x + 3 y = 156.

The manufacturer should produce 21 jackets and 24 pairs of slacks.