G
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Here is another one that has me confused!
3(x+y)=-2y-7
-3y=10-4x
my directions say to use any method to solve.
3(x+y)=-2y-7
-3y=10-4x
my directions say to use any method to solve.
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- Feb 4, 2004
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- 16,582
Multiply out the first equation.
. . . . .3x + 3y = -2y - 7
Then, since you're supposed to use substitution, pick one of the equations, and then pick one of the variables in that equation. (I don't see that either equation or either variable is easier than the other -- I'd have used addition/elimination to solve this, but you don't see to have a choice -- so pick whatever you like.) Solve that equation for that variable. Plug the result into the other equation in place of that variable, and solve.
Eliz.
. . . . .3x + 3y = -2y - 7
Then, since you're supposed to use substitution, pick one of the equations, and then pick one of the variables in that equation. (I don't see that either equation or either variable is easier than the other -- I'd have used addition/elimination to solve this, but you don't see to have a choice -- so pick whatever you like.) Solve that equation for that variable. Plug the result into the other equation in place of that variable, and solve.
Eliz.
Put the first equation in this format:
3x + 3y = -2y - 7
3x + 5y = -7
now you have that, then work on the second equation:
-3y = 10 -4x
4x - 3y = 10
Now:
3x + 5y = -7
4x - 3y = 10
From here I would multiply each equation to get the common factor of 15y and -15y so that they will cancel each other out:
3(3x + 5y = -7) and
5(4x - 3y = 10) to get:
9x + 15y = -21
20x - 15y = 50
that leaves:
29x = 29
x = 1
Now plug this into the original 2nd equation so you get:
-3y=10-4x
-3y = 10 - 4(1)
-3y = 10 - 4
-3y = 6
y = -2
solution is (1, -2)
Hope that helps!