Page 185 - Math Course 2 (Book 1)

Systems of Equations: Substitution

10x = 50 Simplify. Inf nitely Many or No Solutions
10x = 50 Divide each side by 10.
10 10 Example
x = 5 Simplify.
Substitute 5 for x in either equation to f nd the value Use substitution to solve the system of equations.
of y. 2x + 2y = 8
x + y = –2
4x + y = 12 First equation Solve the second equation for y.

4(5) + y = 12 x = 5 x + y = –2 Second equation

20 + y = 12 Simplify. x + y – x = –2 – x Subtract x from each side.

y = –8 Subtract 20 from each side. y = –2 – x Simplify.

y
x Substitute –2 – x for y in the f rst equation.
0 2x + 2y = 8 First equation
4x + y = 12
2x + 2(–2 – x) = 8 y = –x – 2

2x – 4 – 2x = 8 Distributive Property

–4 = 8 Simplify.
(5, –8)
–2x – 3y = 14 The statement –4 = 8 is false. This means there are
no solutions of the system of equations.

The solution is (5, –8).
Answer Answer no solution
The graph verif es the solution.

Your Turn!

Solve Using Substitution

Use substitution to solve the system of equations. Use substitution to solve the system of equations.
y = 2x x + 2y = 1
3x + 4y = 11 5x – 4y = –23
1
A.(1, ) A. (2, –3)
2
B. (1, 2) B. (–1, 1)

1
C. (2, 1) C. (5, )
2
D. (0, 0) D. (–3, 2)

Answer Answer

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