Solve the following system of linear equations by the substitution method. If there is no solution or an infinite number of solutions, so state. Check your answer.
8x + 3y = -16
6x – y = 40
2006-10-02 15:42:52 · 6 answers · asked by Anonymous in Science & Mathematics ➔ Mathematics
8x + 3y = -16
6x – y = 40
Looking at the second equation, then rearrainging, we get:
6x – 40 = y
So now put this into the first equation wherever you see y
8x + 3*( 6x – 40 ) = -16
Now Solve for x.
2006-10-02 15:47:38 · answer #1 · answered by 1,1,2,3,3,4, 5,5,6,6,6, 8,8,8,10 6 · 0⤊ 0⤋
Substitution method involves making one equation the same as the other (but a different sign) so you can cancel them out. In this equation, if we multiply the second equation by 3, then the "y's" will cancel.
So, you will have:
8x + 3y = -16
18x -3y = 120 (See how we multiplied by 3?)
Now add them together just like a regular addition problem.
Add all the x's: 26x
The y's cancel out.
And then the numbers: 104.
So, 26x = 104. Divide to find x. X=4.
Now, plug x into an equation to find y.
8(4) + 3y = -16.
32 + 3y = -16
3y = -48
y = -16
x = 4 and y = -16.
Regards,
Mysstere
2006-10-02 22:50:13 · answer #2 · answered by mysstere 5 · 0⤊ 0⤋
I will start you off. THe second equation can easily be rearranged so that the variable "y" is isolated (alone on one side of the equation).
y = 6x - 40
Now, substitute (6x - 40) into the first equation to replace the variable "y".
So the first equation becomes
8x + 3(6x - 40) = -16
Now expand the bracket, collect like terms, and solve for x.
Once you have x, substitute that value into y = 6x - 40 to solve for y.
Good luck!
2006-10-02 22:47:13 · answer #3 · answered by whatthe 3 · 0⤊ 0⤋
Substitution means get one variable by itself in one equation, then replace it in the other. You can get y by itself pretty easily in equation 2:
6x - y = 40
6x = 40 + y
6x -40 = y
Then replace the y in equation 1 with (6x - 40): 8x + 3(6x - 40) = -16
Distribute the 3: 8x + 18x - 120 = -16
Are you OK from here, I hope so... If not just look above.
2006-10-02 22:48:35 · answer #4 · answered by hayharbr 7 · 0⤊ 0⤋
so take one of the equations and solve for one of the variables. I'l ltake the second one.
6x - y = 40
-y = 40 - 6x
y = 6x - 40
Now plug this solution of y into the first equation.
8x + 3y = -16
8x + 3(6x - 40) = -16
8x + 18x - 120 = -16
26x = 104
x = 4
Now figure out y.
y = 6x - 40
y = 24 - 40
y = -16
THe answer is x = 4, y = -16.
Hope that helps!
2006-10-02 22:48:02 · answer #5 · answered by teekshi33 4 · 0⤊ 0⤋
You need to multiply the 2nd equation by 3 then add that second equation to the first equation to cancel out the y's then you can solve for x.
After you've found x, plug it into the second equation to solve for y.
I hope that was enough help!
2006-10-02 22:45:27 · answer #6 · answered by dualspace 3 · 0⤊ 0⤋