2x+5y=2

3x-2y=3

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• Anonymous

There's 2 ways to solve a system of equations: substitution (solving for 1 variable and plugging it into the other equation) or elimination. In using the elimination method we come up with a way to add or subtract the equations to eliminate a variable. In this case we have to multiply each equation by something in order to get it to "eliminate."

If I multiply the first one by 3 and the second by -2, when I combine the two equations I will have eliminated the x's and just have y's.

3(2x+5y=2)

6x+15y=6

and

-2(3x-2y=3)

-6x+4y=-6

Combine:

6x+15y=6

-6x+4y=-6

_____________

19y=0

y=0

Similarly, multiply the first by 2 and second by 5 to eliminate the y's.

2(2x+5y=2)

4x+10y=4

and

5(3x-2y=3)

15x-10y=15

Combine:

4x+10y=4

15x-10y=15

_____________

19x=19

x=1

And just for fun, we can check it ;)

x=1, y=0

2(1)+5(0)=2

2+0=2

2=2 check :)

3(1)-2(0)=3

3-0+3

3=3 check :)

• Anonymous

You've got 5y in the first and -2y in the second, so multiply the first equation by 2/5 and add it to the second. This gives you

(2/5)2x + 2y + 3x - 2y = 2(2/5) + 3

(2/5)2x + 3x = 2(2/5) + 3

2*2x + 15x = 2(2) + 15

19x = 19

x = 1

Now you can plug this in and solve for y.

• 2x+5y=2 (1)

3x-2y=3 (2)

2 x (1) gives 4x +10y = 4 (3)

5 x (2) gives 15x -10y = 15 (4)

(3) + (4) gives 19x = 19

x = 1

Sub. x=1 into (1)

2 + 5y = 2

5y = 0

y = 0