Question

it would be awesome if u could help me!its about solving systems of equations.

Answer 1

Problem:

Let's denote the above equation like this:

x-3y = -12 EQUATION 1

2x + y = 11 EQUATION 2

Solve x from equation 2:

2x = 11-y

this is equivalent to say (EQUATION 3)

$x\text{ = }\frac{11-y}{2}$

Now, replace the variable x in equation 1, that is:

$(\text{ }\frac{11-y}{2})-\text{ 3y = -12}$

this is equivalent to say:

$\frac{11}{2}-\frac{y}{2}-\text{ 3y = -12}$

Putting together similar terms on different sides of the equation we get:

$\frac{y}{2}+\text{ 3y = }\frac{11}{2}+12$

this is equivalent to:

$\frac{y\text{ + 6y}}{2}\text{= }\frac{11+\text{ 24}}{2}$

this is equivalent to:

$\frac{7\text{y}}{2}\text{= }\frac{35}{2}$

this is equivalent to:

$7\text{y = 35}$

solve for y:

$y\text{ = }\frac{35}{7}\text{ = 5}$

then, we can conclude that y = 5. But, from lines back we know that x is equal to (EQUATION 3):

$x\text{ = }\frac{11-y}{2}$

Replacing y in the previous equation, we obtain

$x\text{ = }\frac{11-y}{2}\text{ = }\frac{11-5}{2}=\text{ }\frac{6}{2}\text{ = 3}$

then, we can conclude that:

X = 3

Y = 5