Question

What’s the answer need it ASAP

Answer 1

Answer:

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

Step-by-step 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.