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.
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.
We will compute A − B.
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.
We will compute F − 2T.
The manufacturer needs to produce 20 sofas and 10 chairs.

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 PreCalculus: Systems of Equations and Inequalities
 PreCalculus: The Slope and Equation of a Line
 PreCalculus: Introduction to Functions
 PreCalculus: Functions and their Graphs
 PreCalculus: Combinations of Functions and Inverse Functions
 PreCalculus: Quadratic Functions
 PreCalculus: Polynomial Functions
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