# Solve the system by substitution

• Apr 10th 2007, 04:35 AM
Patience
Solve the system by substitution
1. 3x + 2y = 11
y = –4x + 8

3x + 2(-4x + 8) = 11
3x – 8x + 16 = 11
3x - 8x + (16 - 16) = 11 - 16
===> -5x = -5
===> x = 1

y = -4(1) + 8
y = -4 + 8
y = 4

so the solution is (1,4) is that right?

Am not sure on equation #2

2. x – 3y = 10
–2x – 4y = 10
• Apr 10th 2007, 04:38 AM
topsquark
Quote:

Originally Posted by Patience
1. 3x + 2y = 11
y = –4x + 8

3x + 2(-4x + 8) = 11
3x – 8x + 16 = 11
3x - 8x + (16 - 16) = 11 - 16
===> -5x = -5
===> x = 1

y = -4(1) + 8
y = -4 + 8
y = 4

so the solution is (1,4) is that right?

This is correct.

1. 3(1) + 2(4) = 11
3 + 8 = 11
11 = 11 (check!)

2. (4)= –4(1) + 8
4 = -4 + 8
4 = 4 (check!)

-Dan
• Apr 10th 2007, 04:40 AM
topsquark
Quote:

Originally Posted by Patience
2. x – 3y = 10
–2x – 4y = 10

In the interest of using a few fractions as possible solve the top equation for x:
x = 3y + 10

and insert this into the bottom equation:
-2(3y + 10) - 4y = 10

-6y - 20 - 4y = 10

-10y - 20 = 10

-10y = 30

y = -3

Thus
x = 3(-3) + 10 = -9 + 10 = 1