What are x and y if `y = 4x + 3` and `2x + 3y = -5`?

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`y = 4x + 3`..........1

`2x + 3y = -5`..........2

put 1 in 2

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

`2x+12x+9=-5`

`14x=-14`

`x=-1`

`y = 4(-1) + 3=-4+3=-1`

There are two ways to algebraically solve for `x` and `y`.

Method 1: Substitution

Through this method, we solve to a variable in one equation and plug it in to the other. In this case, we already know the value of `y` in the first equation. Therefore, we can substitute it it for `y` in the second equation and solve for `x`.

`y=4x+3`
`2x+3(4x+3)=-5`
`2x+12x+9=-5`
`14x=-14`
`x=-1`

Now, we just need to plug `x` back in to one of the equations to solve for `y`. We can use the first equation because `y` is already isolated, but both will yield the same answer.

`y=4(-1)+3)`
`y=-4+3`
`y=-1`

Therefore, `x` is `-1` and `y` is `-1`.

Method 2: Elimination

Through this method, the equations are subtracted so that one of the variables is eliminated. To do this, we must isolate the constant number. In other words, we put `x` and `y` on the same side, like in the second equation.

`y=4x+3`
`0=4x-y+3`
`-3=4x-y`

Now, the equations are both in the same form. However, to eliminate one of the variables, we must get `0` when the equations are subtracted. This means we must have the same coefficients on the variable. For this example, let's solve for `x`. In the first equation, `x` has a coefficient of `4`. Thus, we need `x` in the second equation to have the same coefficient. Because `4` is `2` times its current coefficient of `2`, we need to multiply the entire equation by `2` so it stays equivalent.

`2(2x+3y)=2(-5)`
`4x+6y=-10`

Next, we can subtract the two equations.

`4x+6y=-10`
`-(4x-y=-3)`
–––––––––––––––––––
`0x+7y=-7`

`7y=-7`
`y=-1`

As with the first method, we plug this value back in to find `x`.

`-1=4x+3`
`-4=4x`
`-1=x`

Therefore, `x` is `-1` and `y` is `-1`.