Question

How do you solve the system of equations `4x = 8- 2y` and `5y + 3x = 19`?

Answer 1

See a solution process below:

Explanation:

Solve the first equation for `y`:

`4x = 8 - 2y`

`-color(red)(8) + 4x = -color(red)(8) + 8 - 2y`

`-8 + 4x = 0 - 2y`

`-8 + 4x = -2y`

`(-8 + 4x)/color(red)(-2) = (-2y)/color(red)(-2)`

`(-8)/color(red)(-2) + (4x)/color(red)(-2) = (color(red)(cancel(color(black)(-2)))y)/cancel(color(red)(-2))`

`4 - 2x = y`

`y = 4 - 2x`

Step 2) Substitute `(4 - 2x)` for `y` in the second equation and solve for `x`:

`5y + 3x = 19` becomes:

`5(4 - 2x) + 3x = 19`

`(5 * 4) - (5 * 2x) + 3x = 19`

`20 - 10x + 3x = 19`

`20 + (-10 + 3)x = 19`

`20 + (-7)x = 19`

`20 - 7x = 19`

`-color(red)(20) + 20 - 7x = -color(red)(20) + 19`

`0 - 7x = -1`

`-7x = -1`

`(-7x)/color(red)(-7) = (-1)/color(red)(-7)`

`(color(red)(cancel(color(black)(-7)))x)/cancel(color(red)(-7)) = 1/7`

`x = 1/7`

Step 3) Substitute `1/7` for `x` in the solution to the first equation at the end of Step 1 and calculate `y`:

`y = 4 - 2x` becomes:

`y = 4 - (2 * 1/7)`

`y = 4 - 2/7`

`y = (4 * 7/7) - 2/7`

`y = 28/7 - 2/7`

`y = 26/7`

The Solution Is: `x = 1/7` and `y = 26/7` or `(1/7, 26/7)`

Answer 2

`x=1/7` `&` `y=26/7`

Explanation:

`4x=8-2y`

`=>4xcolor(red)(+2y)=8-2ycolor(red)(+2y)`

`=>bb("4x + 2y= 8")`

Let this be Equation number `1.`

`=>4x+2y=8 -> 1`

`5y+3x=19`

`=>bb("3x + 5y = 19")`

Let this be Equation number `2.`

`=>bb("3x + 5y = 19 ->2")`

Now `bb("Multiply")` Equation number `2` with `2.` That is`,`

`=>3x+5y=19->**2`

`=>(3x+5y)*2=19*2`

`=>3x*2+5y*2= 38`

`=>bb("6x + 10y = 38")`

Let this be Equation number be `3.`

`=>bb("6x + 10y = 38 ->3")`

Now `bb("Multiply")` Equation number `1` with `5.` That is`,`

`=>4x+2y=8->**5`

`=>(4x+2y)*5=8*5`

`=>4x*5+2y*5=40`

`=>bb("20x + 10y = 40")`

Let this be Equation number `4.`

`=>bb("20x + 10y = 40 ->4")`

Now `bb("Subtract")` Equation number `3` from Equation number `4.`

`=>(20x+10y)-(6x+10y)=40-38`

`=>20x+10y-6x-10y=2`

`=>20x-6x+10y-10y=2`

`=>14x=2`

`=>x=2/14`

`=>x=2/(2*7)`

`x=>cancelcolor(red)2/(cancelcolor(red)2*7)`

`x=>1/7`

Now put the value of `x` in Equation number `1.`

That is,

`=> 4*1/7+2y=8`

`=>4/7+2y=8`

`=>2y=8-4/7`

In the Whole number-Fraction Subtraction, `a-b/c=(ac-b)c.` Apply this,

`=>2y=(8*7-4)/7`

`=>2y=(56-4)/7`

`=>2y=52/7`

`=>y=52/(7*2)`

`=>y=(2*26)/(2*7)`

`=>y=(cancelcolor(red)2*26)/(cancelcolor(red)2*7)`

`=>y = 26/7`

So, `x=1/7` `&` `y=26/7`