126k views
4 votes
A student 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

Please help Im stuck.

User AnkeyNigam
by
4.8k points

1 Answer

4 votes

Answer:

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)}....

User Scott Kuhl
by
5.4k points
Welcome to QAmmunity.org, where you can ask questions and receive answers from other members of our community.