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

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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)...Show more

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...Show more

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...Show more