How do you solve an equotion, with 2 variables? |
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 Oct 27 2005, 06:15 PM
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#1
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 Senior Member  Group: Member Posts: 103 Joined: Jul 2004 Member No: 31,102  |
Ah for some reason I can't remember.
I know you can substitute it but I dont' know if I'm doing it right. It's a equiangular triangle and they give the measures of the sides and I know they all equal each other I just can't get x & y 4x -y=7 2x+3y=7 Then I get kinda stuck haha.. can anyone help? You don't have to tell me the answer just like an example or whatever |
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| *mipadi* |
Oct 27 2005, 08:08 PM
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#2
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Choose a variable (let's say x). Now, use one equation to solve for y in terms of x, then subtitute that into the second equation. For example:
4x - y = -7 -y = -7 - 4x y = 7 + 4x 2x + 3(7 + 4x) = 7 2x + 21 + 12x = 7 14x + 21 = 7 Now, just solve like normal. |
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Oct 27 2005, 10:12 PM
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#3
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 Peggy. Group: Member Posts: 2,508 Joined: Aug 2005 Member No: 214,025 |
Substitution or elimination.
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Oct 28 2005, 07:00 AM
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#4
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 hi. call me linda. Group: Official Member Posts: 8,187 Joined: Feb 2004 Member No: 3,475 |
You could also multiply the first equation by 3, and get:
12x - 3y = 21. Then, add the second equation to the first so the y cancels out: 12x - 3y = 21 - 2x + 3y = 7 ------------------ 10x = 14 Divide by 10 and you get x=7/5. Then subsitute that into one of the equations and solve for y. |
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Oct 28 2005, 02:45 PM
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#5
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Senior Member Group: Member Posts: 103 Joined: Jul 2004 Member No: 31,102 |
Okay thanks. this can be closed now
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| *swtcherriipie* |
Oct 28 2005, 02:49 PM
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#6
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I havent done that in 2 weeks dont remind me calculus SUCKS!
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| *mipadi* |
Oct 28 2005, 03:42 PM
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#7
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QUOTE(swtcherriipie @ Oct 28 2005, 3:49 PM) I havent done that in 2 weeks dont remind me calculus SUCKS!  That's not calculus, it's algebra. 
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Oct 28 2005, 03:47 PM
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#8
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 Senior Member Group: Member Posts: 59 Joined: Oct 2005 Member No: 264,617 |
X is 2 and Y is 1. Yes sirree.
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Oct 29 2005, 12:09 AM
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#9
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 Senior Member Group: Staff Alumni Posts: 4,799 Joined: Aug 2004 Member No: 37,450 |
QUOTE(doubleaxel1951 @ Oct 28 2005, 2:45 PM) Okay thanks. this can be closed now Topic Closed |
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