Question

What’s the answer need it ASAP

Answer 1

The pair of solutions are
`(-2,-1) \ and\ (3,14)`

Explanation:

`1)y=x^2+2x-1\\2)y-3x=5`

Solving for
`y` in equation 2.

`y-3x=5`

Adding
`3x` both sides.

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

`y=5+3x`

Substituting value of
`y` in equation 1.

`5+3x=x^2+2x-1\\`

Simplifying it further. Subtracting
`5` from both sides.

`5+3x-5=x^2+2x-1-5\\`

`3x=x^2+2x-6\\`

Subtracting
`3x` both sides.

`3x-3x=x^2+2x-6-3x\\`

`0=x^2-x-6\\`

Now we have a quadratic equation to solve.

`x^2-x-6=0`

Solving quadratic by factor method.

Splitting the middle term into two terms
`mx \ and\ nx` such that
`mx+nx=-x \ and (mx)(nx)=-6x^2`

By factoring 6 we can get the terms.

`mx=-2x \ and\ nx=3x` As we know
`[2x-3x=-x \ and\ (2x)(-3x)=-6x^2]`

So, the equation would be

`x^2+2x-3x-6=0`

Now, we factor in pairs.

Taking
`x` as common factor from first two terms and taking
`-3` common from last two terms.

`x(x+2)-3(x+2)=0`

Taking
`(x+2)` as common factor from the whole.

`(x+2)(x-3)=0`

The roots can be written as:

`x+2=0 \ and\ x-3=0`

Solving for
`x` from the above we get

`x=-2 \ and\ x=3`

Plugging the values of
`x` in the equation
`y=5+3x`

When
`x=-2`

`y=5+3(-2)`

`y=5-6`

`y=-1`

So
`(-2,-1)` is one solution.

When
`x=3`

`y=5+3(3)`

`y=5+9`

`y=14`

So
`(3,14)` is another solution.