Substitution

Systems of linear equations can be solved using the substitution method, in which the values of variables are substituted into equations in order to solve one variable at a time.

This method is often used when one of the variables of an equation has no coefficient, as x is shown in equation A.	Original Equations: A.) x +2y = 2 B.) 3x + 4y = -4
1.) The first step is to solve one of the equations for one variable. Since x in equation A is without a coefficient, first solve for x by rearranging the equation as shown to the right.	A.) x = -2y + 2
2.) Next, substitute the expression from step 1 into equation B and solve for y.	B.) 3x + 4y = -4 B.) 3(-2y + 2) + 4y = -4 y = 5
3.) Now substitute the value from step 2 into equation A and solve for x.	A.) x = -2y + 2 A.) x = -2(5) + 2 x = -8
The solution to this system of equations is written as (-8, 5).

1.) The problem: The admission fee at a small fair is $1.50 for children and $4.00 for adults. On a certain day, 2200 people enter the fair and $5050 is collected. How many children and adults attended?	1.a) Convert the problem into equations: A.) a + c = 2200 B.) 4a + 1.5c = 5050
2.) Solve the system for the number of adults.	2.a) spaceA.) a = 2200 - c
3.) Substitute the value of a into equation B and solve for c.	3.a) spacB.) 4a + 1.5c = 5050 B.) 4(2200 - c) + 1.5c = 5050 8800 - 4c + 1.5c = 5050 8800 - 2.5c = 5050 -2.5c = -3750 c = 1500
4.) Now substitute the value of c into equation A and solve for a.	4.a) spacA.) a + c = 2200 A.) a + (1500) = 2200 a = 700
The solution: 1500 children and 700 adults attended.