# How to solve systems of equations with multiplication & addition.

How to solve systems of equations with multiplication & addition.
$a+b=12$
$a\cdot b=36$
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Alexia Hart
There are two ways to solve a system of equations, the first one being substitution and other one being elimination. In this case, we will use substitution which requires us to put one variable in terms of another. Currently we have the following equations:
$\begin{array}{rl}& a+b=12\\ & a\cdot b=36\end{array}$
Since we have two variables and as I said before, we have to rearrange one in terms of another. We can either choose $a$ or $b$. We'll choose $a$. If rearrange for $a$ we get:
$a=-b+12$
Since,we have said that $a=b-12$, we can substitute $a$ into our second equation so we get:
$\left(-b+12\right)\left(b\right)=36$
Now we can expand and see that we have a quadratic equation:
$-{b}^{2}+12b=36$
Now we bring $36$ to the left side and make equation equal to $0$.
$-{b}^{2}-12b-36=0$
There is no need to get complex here but for future equations, you should check if the equation has $1$, $2$, or no real roots through the discriminant ${b}^{2}-4ac$. If the value is below zero then it has no real roots, if it is above zero than it has $2$ real roots and if it is equal to zero then it has $1$ root. In this case, the discriminant is equal to zero and therefore, has only one root.
You can use whatever method you like to solve equation but the answer that you will get is $6$.
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civilnogwu
If you are given numbers $r$ and $s$, and you have to find numbers $a$ and $b$ such that
$a+b=r,\phantom{\rule{2em}{0ex}}ab=s$
then you can think about the polynomial
$\left(x-a\right)\left(x-b\right)={x}^{2}-\left(a+b\right)x+ab={x}^{2}-rx+s$
So, the solutions of the quadratic equation
${x}^{2}-rx+s=0$
are the numbers $a$ and $b$ that you want. Can you solve a quadratic equation?