3x+4y=15 and 2x-4y=30 Solving systems of Equations using Substitution and elimination

Question

Some Algebra 1 ive need some help on.

Answer 1

line 1: 3x+4y=15

line 2: 2x-4y=30    or 4y=2x-30

line1 +line2...(3x+2x)+(4y-4y)=45

5x=45

x=9

y=(1/2)x -15/2

=(9/2)-(15/2)=-6/2 =-3

Answer 2

3x+4y=15 ( 1 ) and 2x-4y=30 ( 2 )

By Substitution:

Solve for x :

y = ( 15 - 3x ) / 4    ( 1 )

Substitute to ( 2 ):

2x - 4 ( ( 15 - 3x ) / 4 ) = 30

2x - ( 60 - 12x ) / 4 = 30

( 8x - 60 + 12x ) / 4 = 30  ( cross multiply )

20x - 60 = 120

20x = 180

x = 9  , y = 3

 

By Elimination:

3x+4y=15 ( 1 )

2x-4y=30 ( 2 )

First, eliminate y to get x by adding the two equations and the result would be:

5x = 45

x = 9

Eliminate x to get y by multiplying ( 1 ) by 2 and ( 2 ) by 3, and the result is:

6x + 8y = 30 ( 3 )

6x - 12y = 90 ( 4 )

Substract ( 3 ) by ( 4 ) to get the value of y, and the result would be:

20y = - 60

y = - 3.

 

Answer 3

3x+4y = 15 -----> (1)

2x-4y = 30 --------> (2)

To eliminate the y value add the equations (1) and(2).

3x+4y = 15

2x-4y = 30

_________

5x = 45

Divide to each side by 5.

5x/5 = 45/5

x = 9

Substitute the x value in (1).

3*9+4y = 15

27+4y = 15

Subtract 27 from each side.

27-27+4y = 15-27

4y = -12

Divide to each side by 4.

4y/4 = -12/4

y = - 3

Solution is x =9, y = -3