Solving Systems of Equations Using Substitution

Examples

Example 1:

Solve the second equation for x and use the expression as a replacement for x in the first equation, which gives us an equation in y alone.

→ x = y + 1 Replace for x in the equation [1]: 2(y + 1) + y = 5

Solve this equation for y:

Replace y = 1 in equation [2]: x = ( 1 ) + 1 and find x = 2

Always check in the other equation: 2(2) + (1) = 5

The solution is the point (2, 1) or S = { ( 2, 1) } y = -x + 8 3x − 2y = -1 2x + y = 5 x − y = 1

Example 2:

Solve the second equation for y : y = 3x − 1 use the expression on the right side as a replacement for y in the first equation, which gives us an equation in x alone.

Substitute the expression for y in the second equation: 2x + 3(3x − 1) = 19

Solve this equation for x:

Replace in equation [2]: y = 3 (2 ) − 1 or y = 5

Always check by replacing both in first equation: 2(2) + 3 (5) = 19

Thus, the solution is the point (2, 5): or S = { ( 2, 5) }

Example 3:

Treat decimal fractions like common fractions. Multiply by LCM.

Multiply equation [1] by LCM =100:

Solve the second equation for x and use the expression x = 27 − y

as a replacement for x in the new first equation: 5(27 − y) + 10y = 235

Solve this equation for y:

Replace y = 20 in equation [2]: x = 27 − ( 20 ) and find x = 7

Always check in the other equation: .05(7) + .10(20) = 2.35

The solution is the point (7, 20) or S = { ( 7, 20) }