Answer:
The pair of solutions are [tex](-2,-1) \ and\ (3,14)[/tex]
Step-by-step explanation:
[tex]1)y=x^2+2x-1\\2)y-3x=5[/tex]
Solving for [tex]y[/tex] in equation 2.
[tex]y-3x=5[/tex]
Adding [tex]3x[/tex] both sides.
[tex]y-3x+3x=5+3x[/tex]
[tex]y=5+3x[/tex]
Substituting value of [tex]y[/tex] in equation 1.
[tex]5+3x=x^2+2x-1\\[/tex]
Simplifying it further. Subtracting [tex]5[/tex] from both sides.
[tex]5+3x-5=x^2+2x-1-5\\[/tex]
[tex]3x=x^2+2x-6\\[/tex]
Subtracting [tex]3x[/tex] both sides.
[tex]3x-3x=x^2+2x-6-3x\\[/tex]
[tex]0=x^2-x-6\\[/tex]
Now we have a quadratic equation to solve.
[tex]x^2-x-6=0[/tex]
Solving quadratic by factor method.
Splitting the middle term into two terms [tex]mx \ and\ nx[/tex] such that [tex]mx+nx=-x \ and (mx)(nx)=-6x^2[/tex]
By factoring 6 we can get the terms.
[tex]mx=-2x \ and\ nx=3x[/tex] As we know [tex][2x-3x=-x \ and\ (2x)(-3x)=-6x^2][/tex]
So, the equation would be
[tex]x^2+2x-3x-6=0[/tex]
Now, we factor in pairs.
Taking [tex]x[/tex] as common factor from first two terms and taking [tex]-3[/tex] common from last two terms.
[tex]x(x+2)-3(x+2)=0[/tex]
Taking [tex](x+2)[/tex] as common factor from the whole.
[tex](x+2)(x-3)=0[/tex]
The roots can be written as:
[tex]x+2=0 \ and\ x-3=0[/tex]
Solving for [tex]x[/tex] from the above we get
[tex]x=-2 \ and\ x=3[/tex]
Plugging the values of [tex]x[/tex] in the equation [tex]y=5+3x[/tex]
When [tex]x=-2[/tex]
[tex]y=5+3(-2)[/tex]
[tex]y=5-6[/tex]
[tex]y=-1[/tex]
So [tex](-2,-1)[/tex] is one solution.
When [tex]x=3[/tex]
[tex]y=5+3(3)[/tex]
[tex]y=5+9[/tex]
[tex]y=14[/tex]
So [tex](3,14)[/tex] is another solution.
