Astudent likes to use the substitution method for systems of equation. how can he use it with a system that is not in the proper form for substitution?
show with this system:
-2x+y=4
3x+4y=49
im stuck.

Step-by-step explanation:

We have given:

-2x+y=4     equation1

3x+4y=49 equation 2

We will solve the 1st equation for y and substitute the value into the 2nd equation.

-2x+y=4     equation1

Move the values to the R.H.S except y

y = 2x+4

Now substitute the value of y in  2nd equation:

3x+4y=49

3x+4(2x+4)=49

3x+8x+16=49

Combine the like terms:

3x+8x=49-16

11x=33

Now divide both the sides by 11

11x/11 = 33/11

x= 3

Now substitute the value of x in any of the above equations: We will substitute the value in equation 1:

-2x+y=4

-2(3)+y=4

-6+y=4

Combine the constants:

y=4+6

y = 10

Thus the solution set of (x,y) is {(3,10)}

idk

step-by-step explanation:

your victory is right around the corner. never give up.

- niki minaj