solve the elimination method for
3x+7y=-32
7x+3y=-8
10x+10y=40
dived by 10
x=4
1st times 3 --->9x + 21y = -96
2nd times 7 --> 49x + 21y = -56
subtract them:
40x = 40
x = 1
sub into 1st:
3 + 7y = -32
7y = -35
y = -5
so x = 1 and y = -5
To solve the system of equations using the elimination method, you need to eliminate one variable by multiplying one or both equations by appropriate coefficients so that when you add or subtract them, one variable is eliminated.
In this case, multiplying the first equation by 7 and the second equation by 3 will allow us to eliminate the variable "x". Remember that when you multiply an equation by a number, you have to multiply each term in the equation by that number.
Multiply the first equation by 7:
(7)(3x + 7y) = (7)(-32)
21x + 49y = -224 <- Equation A
Multiply the second equation by 3:
(3)(7x + 3y) = (3)(-8)
21x + 9y = -24 <- Equation B
Now, subtract Equation B from Equation A to eliminate the "x" variable:
(21x + 49y) - (21x + 9y) = (-224) - (-24)
21x - 21x + 49y - 9y = -224 + 24
0x + 40y = -200
40y = -200
To solve for "y", divide both sides of the equation by 40:
40y/40 = -200/40
y = -5
Now that we have the value of "y", we can substitute it back into one of the original equations to solve for "x". Let's use the first equation:
3x + 7(-5) = -32
3x - 35 = -32
Next, isolate the "x" term by adding 35 to both sides of the equation:
3x - 35 + 35 = -32 + 35
3x = 3
Finally, solve for "x" by dividing both sides of the equation by 3:
3x/3 = 3/3
x = 1
So the solution to the system of equations is x = 1 and y = -5.