We have learned how to solve many kinds of systems of equations using algebra that would be difficult to solve by graphing. For example, look at

\(\begin{cases} y = 2x -3 \\ x+2y=7 \end{cases}\)

The first equation says that $y=2x-3$, so wherever we see $y$, we can substitute the expression $2x-3$ instead. So the second equation becomes $x+2(2x-3) = 7$.

We can then solve for $x$:

\(\begin{align} x + 4x-6 &= 7 && \text{distributive property} \\ 5x-6 &= 7 && \text{combine like terms} \\ 5x &= 13 && \text{add 6 to each side} \\ x &= \frac{13}{5} && \text{multiply each side by $\frac15$} \end{align}\)

We know that the $y$ value for the solution is the same for either equation, so we can use either equation to solve for it. Using the first equation, we get:

\(\begin{align} y &= 2\left(\frac{13}{5}\right)-3 &&\text{substitute $x=\frac{13}{5}$ into the equation}\\ y &= \frac{26}{5}-3 &&\text{multiply $2\left(\frac{13}{5}\right)$ to make $\frac{26}5$}\\ y &=\frac{26}{5} - \frac{15}5 &&\text{rewrite 3 as $\frac{15}5$}\\ y &= \frac{11}5 \end{align}\)

If we substitute $x=\frac{13}5$ into the other equation, $x+2y=7$, we get the same $y$ value. So the solution to the system is $\left(\frac{13}{5},\frac{11}5\right)$.

There are many kinds of systems of equations that we will learn how to solve in future grades, like \(\begin{cases} 2x+3y = 6 \\ \text-x+2y = 3 \end{cases}\).

Or even \(\begin{cases} y = x^2 +1 \\ y = 2x+3 \end{cases}\).