How do you solve this system of equations: #-x + 2y = 10 and 3x + 4y = - 10#?

Question

Genesis Allen · Accepted Answer

the solutions are #x=-6,y=2# given system of equations #-x+2y=10->(1)# #3x+4y=-10->(2)# multiplying# (1)# by #3# we get #-3x+6y=30->(3)# #(3)+(4)# #-3x+6y-30+3x+4y+10=0rArr10y-20=0# #rArr10y=20rArry=2# put #y=2 # in #(3)# we get #-3x+12=30rArr-3x=18rArrx=-6# #:.x=-6,y=2#

Jameson Allen · Answer

undefined To solve the system of equations -x + 2y = 10 and 3x + 4y = -10, you can use either the substitution method or the elimination method.

Using the elimination method, you can multiply the first equation by 3 and the second equation by 1 to make the coefficients of x opposite. This will allow you to eliminate x when you add the equations together.

The equations become:
-3x + 6y = 30
3x + 4y = -10

Adding the equations, you get:
10y = 20

Dividing both sides by 10, you find:
y = 2

Substituting y = 2 into either of the original equations, such as the first one, you can solve for x:
-x + 2(2) = 10
-x + 4 = 10
-x = 6
x = -6

So, the solution to the system of equations is x = -6 and y = 2.