SOLUTION: How would I solve these systems? 1)x+5y=4 3x-7y=-10 2)-5x+3y=6 x-y=4 3)3x-8y=13 4x-5y=6 4)3x-7y=6 2x+7y=4 5)4x+3y=19 3x-4y=8

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Question 22279: How would I solve these systems?
1)x+5y=4
3x-7y=-10

2)-5x+3y=6
x-y=4

3)3x-8y=13
4x-5y=6

4)3x-7y=6
2x+7y=4

5)4x+3y=19
3x-4y=8

6)2x+y=5
3x-2y=4

7)3x-2y=19
5x+4y=17

8)7x+3y=-1
4x+y=3

9)8x-3y=-11
2x-5y=27

10)4x-7y=10
3x+2y=-7

Answer by elima(1433)   (Show Source): You can put this solution on YOUR website!
There are different ways to solve thewe, one way is by substitution; where one equation is solved t\for one variable and that solutin is substituted into the second equation. I will solve a few, then you try and do the rest.
Solve one equation for one variable;
1)x+5y=4 (eq.1)
3x-7y=-10 (eq.2)
Lets start by solving eq.1 for x;
x=4-5y (eq.3)
Now substitute eq.3 into the other equation and solve for y;
3(4-5y)-7y=-10
12-15y-7y=-10
12-22y=-10
-22y=-10-22
-22y=-22
y=1
Now you have what y equals, so you are going to plug that into eq.1 and solve for x;
x+5(1)=4
x+5=4
x=4-5
x=-1
So the answer; x=-1, y=1
2)-5x+3y=6 (eq.1)
x-y=4 (eq.2)
x-y=4
x=4+y (eq.3)
-5(4+y)+3y=6
-20-5y+3y=6
-20-2y=6
-2y=6+20
-2y=26
y=-13
x-y=4 (eq.2)
x-(-13)=4
x+13=4
x=4-13
x=-9
x=-9, y=-13
Now try and solve the others on your own, if you are still having problems, let me know.
=)


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