Asked by diana
Solve the following systems of equations.
3x + 4y = 4
2x + y = 6
3x + 4y = 4
2x + y = 6
Answers
Answered by
Bosnian
2 x + y = 6 Subtract 2 x to both sides
2 x + y - 2 x = 6 - 2x
y = 6 - 2x
3 x + 4 y = 4
3 x + 4 ( 6 - 2 x ) = 4
3 x + 24 - 8 x = 4
- 5 x + 24 = 4 Subtract 24 to both sides
- 5 x + 24 - 24 = 4 - 24
- 5 x = - 20 Divide both sides by - 5
- 5 x / - 5 = - 20 / - 5
x = 4
y = 6 - 2x
y = 6 - 2 * 4
y = 6 - 8 = - 2
Solution :
x = 4 , y = - 2
2 x + y - 2 x = 6 - 2x
y = 6 - 2x
3 x + 4 y = 4
3 x + 4 ( 6 - 2 x ) = 4
3 x + 24 - 8 x = 4
- 5 x + 24 = 4 Subtract 24 to both sides
- 5 x + 24 - 24 = 4 - 24
- 5 x = - 20 Divide both sides by - 5
- 5 x / - 5 = - 20 / - 5
x = 4
y = 6 - 2x
y = 6 - 2 * 4
y = 6 - 8 = - 2
Solution :
x = 4 , y = - 2
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