Page 193 - Math Course 2 (Book 1)

Solving Systems of Equations by Multiplying

Mo. 6

Lesson 4 Multiply Both Equations to
Eliminate

KEY CONCEPTS: Example
1. Solve systems of equations algebraically
by using elimination with multiplication. Use elimination to solve the system of equations.
2. Solve real-world problems involving 4x + 3y = 8
systems of equations. 3x – 5y = –23
Method 1 Eliminate x.
4x + 3y = 8 12x + 9y = 24
3x – 5y = –23 (+)–12x + 20y = 92
MO. 6 - L4a
12x + 9y = 24 Multiply by 3.
Solving Equations: (+)–12x + 20y = 92 Multiply by –4.
Elimination with Multiplication 29y = 116 Add the equations.

29y = 166 Divide each side by 29.
29 29
Let’s Begin y = 4 Simplify.

Now substitute 4 for y in either equation to f nd x.

Multiply One Equation to Eliminate 4x + 3y = 8 First equation
4x + 3(4) = 8 y = 4

Example 4x + 12 = 8 Simplify.
4x + 12 – 12 = 8 – 12 Subtract 12 from each side.
4x –4
Use elimination to solve the system of equations. 4 = 4 Divide each side by 4.
2x + y = 23 x = –1 Simplify.
3x + 2y = 37

Multiply the f rst equation by –2 so the coef cients The solution is (–1, 4)
of the y terms are additive inverses. Then add the Answer
equations.

2x + y = 23 –4x – 2y = –46 Multiply by –2.
3x + 2y = 37 (+) 3x + 2y = 37 Method 2 Eliminate y.
–x = –9 Add the equations. 4x + 3y = 8 20x + 15y = 40
3x – 5y = –23 (+) 9x – 15y = –69
–x = –9 Divide each side
–1 –1 by –1. 20x + 15y = 40 Multiply by 5.
(+) 9x – 15y = –69 Multiply by 3.
x = 9 Simplify.
Now substitute 9 for x in either equation to f nd the 29x = –29 Add the equations.
value of y.
2x + y = 23 First equation 29x = –29 Divide each side by 29.
2(9) + y = 23 x = 9 29 29
18 + y = 23 Simplify.
x = –1 Simplify.
18 + y – 18 = 23 – 18 Subtract 18 from each side.
y = 5 Simplify.

Answer The solution is (9, 5)

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