PLEASE HELP how do i solve the system of equation algebraically and check?

First equation: solve for x

5x-y=4

+y +y

5x=4+y

/5. /5

X=4/5 +y

Second equation: insurer the Value of x (found in 1st equation)

4x+2y=6

4(4/5+y)+2y=6

3.2+4y+2y=6

3.2+6y=6

-3.2 -3.2

6y=2.8

/6. /6

Y=.46

Substitute y in in first equation

5x-y=4

5x-.46=4

+.46. +.46

5x=4.46

/5. /5

x=.90

Ordered pair

(.90,.46)

you need to make one of the coefficients disappear by multiplication and algebraic addition

Lets choose the first equation and increase -y to -2y by multiplying the entire equation by 2

i.e it becomes 10x-2y=8

now algebraic addition

10x-2y=8

4x+2y=6

--------------

14x+0y=14

i.e x=1 now substitute the value of x into one of the equations

5x-y=4 becomes 5(1)-y=4 becomes -y=-1 i.e y=1

5x-y=4

-4

5x-y-4

+y

5x-4=y

and then...

5x-y=4 (beginning again, solve for x)

+y

5x= y+4

/5

x= (y+4) divided by 5

now do the same for the other one