Use the elimination method to solve the system of equations

5u+2v=15
3u+v=-7

1st one: 5u + 2v = 15

double the 2nd: 6u + 2v = -14

subtract them:
-u = 29
u = -29

sub into 2nd:
3(-29) + v = -7
v= -7 + 87 = 80

u = -29 , v = 80

To solve the system of equations using the elimination method, we need to eliminate one of the variables by multiplying one or both equations by appropriate numbers. The goal is to create a new system of equations where the coefficient of one variable in one equation cancels out with the coefficient of the same variable in the other equation.

Let's start by multiplying the second equation by 2 to eliminate the variable "v" and make the coefficients of "v" in both equations the same:

Original equations:
5u + 2v = 15 ...(1)
3u + v = -7 ...(2)

Multiplying equation (2) by 2:
2(3u + v) = 2(-7)
6u + 2v = -14 ...(3)

Now we have a system of equations where the coefficients of "v" are the same in equations (1) and (3):

5u + 2v = 15 ...(1)
6u + 2v = -14 ...(3)

Now we can subtract equation (1) from equation (3) to eliminate "v":

(6u + 2v) - (5u + 2v) = (-14) - (15)
6u - 5u + 2v - 2v = -14 - 15
u = -29

We have found the value of "u" to be -29.

To find the value of "v," we substitute the value we found for "u" back into one of the original equations. Let's use equation (1):

5u + 2v = 15

Substituting u = -29:

5(-29) + 2v = 15
-145 + 2v = 15
2v = 15 + 145
2v = 160
v = 160 / 2
v = 80

We have found the value of "v" to be 80.

Therefore, the solution to the system of equations is u = -29 and v = 80.