SOLUTION: 2y = x + 3 5y = x - 7 What is the solution set of the given system? {(-29/3, -10/3)} {(29/3, 10/3)} {(1/3, 4/3)} {(-1/3, -4/3)}

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Question 1170800: 2y = x + 3
5y = x - 7
What is the solution set of the given system?
{(-29/3, -10/3)}
{(29/3, 10/3)}
{(1/3, 4/3)}
{(-1/3, -4/3)}
Found 3 solutions by Cromlix, Alan3354, MathTherapy:
Answer by Cromlix(4381)   (Show Source): You can put this solution on YOUR website!
Hi there,
2y = x + 3
5y = x - 7
Rearrange
2y - x = 3......Eq(1)
5y - x = -7.....Eq(2)
Multiply Eq(2) by -1
2y - x = 3......Eq(1)
-5y + x = 7.....Eq(2)
Add Eq(1) + Eq(2)
-3y = 10
y = -10/3
Substitute y = -10/3 into Eq(1)
2y - x = 3......Eq(1)
2(-10/3) - x = 3
-20/3 - x = 3
-x = 3 + 20/3
3 = 9/3
-x = 9/3 + 20/3
x = - 29/3
{(-29/3, - 10/3)}
Hope this helps :-)
Answer by Alan3354(69443)   (Show Source): You can put this solution on YOUR website!
2y = x + 3
5y = x - 7
--------------------------- Subtract
-3y = 10
y = -10/3
---
Sub for y in either eqn and find x
Answer by MathTherapy(10552)   (Show Source): You can put this solution on YOUR website!

2y = x + 3
5y = x - 7
What is the solution set of the given system?
{(-29/3, -10/3)}
{(29/3, 10/3)}
{(1/3, 4/3)}
{(-1/3, -4/3)}
2y = x + 3 ---- eq (i)

5y = x - 7 ---- eq (ii)

3y = - 10 ----- Subtracting eq (i) from eq (ii)





 ------- Substituting  for y in eq (i)

 

Solution set:  

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