SOLUTION: I知 very confused about solving this problem and its supposed to be solved using substitution. The linear system solve it for me but I need the steps. Thanks 10x+8y=5 8x+6y=3

Algebra ->  Coordinate Systems and Linear Equations -> SOLUTION: I知 very confused about solving this problem and its supposed to be solved using substitution. The linear system solve it for me but I need the steps. Thanks 10x+8y=5 8x+6y=3       Log On


   

Question 13377: I知 very confused about solving this problem and its supposed to be solved using substitution. The linear system solve it for me but I need the steps. Thanks
10x+8y=5
8x+6y=3
Answer by akmb1215(68) About Me  (Show Source):
You can put this solution on YOUR website!
The easiest way to solve this system of equations is by combination. You need to figure out the least common multiple of 10 and 8. This is 40. To get 40 from 10, you multiply by 4...so multiply the entire top equation by 4 to get 40x+%2B+32y+=+20. In order to get rid of the x in the combination, you need to get -40x in the second equation. To get this, multiply the entire equation by -5 to get -40x+-+30y+=+-15. Now, you can add the two equations together to get...2y+=+5. Solve for y to get 5%2F2+=+y. To get x, plug this into either equation (I am going to use the top one). I get 10x+%2B+8%282.5%29+=+5. Multilply to get 10x+%2B+20+=+5. Move 20 to the right side of the equation to get 10x+=+-15. Divide to get x+=+-15%2F10, which reduces to x+=+-3%2F2.