Question

What is the solution to this system of linear equations?

Answer 1

Answer:  The required solution to the given system of equation is (x, y) = (-8, -2).

Step-by-step explanation:  We are given to find the solution of the following system  of equations :

$y-x=6~~~~~~~~~~~~~~~~~~~~~~~~~~(i)\\\\y+x=-10~~~~~~~~~~~~~~~~~~~~~~(ii)$

Adding equations (i) and (ii), we have

$(y-x)+(y+x)=6+(-10)\\\\\Rightarrow 2y=6-10\\\\\Rightarrow 2y=-4\\\\\Rightarrow y=\dfrac{-4}{2}\\\\\Rightarrow y=-2.$

Substituting the value of y in equation (i), we get

$y-x=6\\\\\Rightarrow x=y-6\\\\\Rightarrow x=-2-6\\\\\Rightarrow x=-8.$

Thus, the required solution to the given system of equation is (x, y) = (-8, -2).

Answer 2

To solve for any given number of variables, you need that number of equations to solve for all variables.  When there are only two variables you only need to equations to solve for them both.  You eliminate one variable at a time, order does not matter.

y-x=6 and y+x=-10 

Since one equation has -x and the other +x, you can simply add the two equations together to eliminate x...

(y-x=6)+(y+x=-10)

y+y-x+x=6-10

2y=-4

y=-2, now you can use this value of y in either original equation to solve for x...

y-x=6, using y=-2 becomes:

-2-x=6

-x=8

x=-8

So the solution to the system of equations is the point (-8, -2)

You can always check by using these two values in the original equations...

y-x=6, -2--8=6, -2+8=6, 6=6

y+x=-10, -2+(-8)=-10, -2-8=-10, -10=-10