Hi, I am really stuck with this equations. I am so lost and dont know what to do. Please help. x - 2 + y - 1 = 13 (eq.1) 3 4 12 2 - x + 3 + y = 11 (eq.2) 2 3 6
Last edited by NettyJay; Nov 17th 2006 at 04:48 AM. Reason: it didnt come out the way i typed it
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Originally Posted by NettyJay Hi, I am really stuck with this equations. I am so lost and dont know what to do. Please help. x - 2 + y - 1 = 13 (eq.1) 3 4 12 2 - x + 3 + y = 11 (eq.2) 2 3 6 Did you mean: Originally Posted by NettyJay Hi, I am really stuck with this equations. I am so lost and dont know what to do. Please help. (eq.1) (eq.2)
Originally Posted by NettyJay Hi, I am really stuck with this equations. I am so lost and dont know what to do. Please help. Hi, I am really stuck with this equations. I am so lost and dont know what to do. Please help. (eq.1) (eq.2) I'll do you try the other one. First thing to do is get rid of denominators, so multiply everything by 3 to get: Now multiply by 4: Simplify: Thus: Add 11 to both sides: Subtract 3y from both sides: Then divide by 4: Solve: That can be simplified to:
Was this supposed to be a system of equations, meaning was I sposed to find the value of x and y to fit both of them?
Originally Posted by NettyJay Hi, I am really stuck with this equations. I am so lost and dont know what to do. Please help. x - 2 + y - 1 = 13 (eq.1) ...3....... 4..... 12 2 - x + 3 + y = 11 (eq.2) ..2........ 3....... 6 Multiply the first equation through by to get: rearrange to: Multiply the second equation through by to get: rearrange: So you have a bog standard pair of simultaneous linear equations: ... RonL
Just to extend CaptainBlack's post. You've got system of equations to solve: Solve first equation for : Substitute into second equation: Now we can find from first equation by substituting : So solutions are
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