Solution

First rearrange the equations to make them easier to work with.

\begin{align} 4y -2x &= 7 &\color{purple}{(\star)}\\ 2y +10x &= 20 &\color{purple}{(\dagger)}\\ \end{align}

To find the point of intersection, you multiply one of the equations by a number. This number is chosen so that one of the following is true:

A. a term (e.g. $10x$) appears in both equations.

OR

B. a term is positive in one of the equations (e.g. $10x$) and negative ($-10x$) in the other.

So, if you multiply $\color{purple}{(\star)}$ by $5$, we then have:

\begin{align} 20y -10x &= 35 &\color{purple}{(\star \star)}\\ 2y +10x &= 20. &\color{purple}{(\dagger \dagger)}\\ \end{align}

Here, one equation contains $-10x$ and the other has $10x$, (case B). (Note: Here, $\color{purple}{(\dagger \dagger)}$ is the same as $\color{purple}{(\dagger)}$.)

On the other hand, you could multiply $\color{purple}{(\dagger)}$ by $2$, which would result in:

\begin{align} 4y -2x &= 7 &\color{purple}{(\star \star \star)}\\ 4y +20x &= 40. &\color{purple}{(\dagger \dagger \dagger)}\\ \end{align}

Here both equations contain the term $4y$, (case A). (Note: $\color{purple}{(\star \star \star)}$ is the same as $\color{purple}{(\star)}$ here.)

For case A, we subtract one equation from the other, term by term:

\begin{align} &4y -2x=7\\ &4y +20x= 40\\ &\overline{\quad\;\, -22x=-33} \end{align}

We can now solve $-22x=-33$ by dividing both sides by $-22$ to get $x = \dfrac{3}{2}$. Now substitute our solution for $x$ into either of $\color{purple}{(\dagger)}$ or $\color{purple}{(\star)}$ to find $y$. Substituting $x=\dfrac{3}{2}$ into $\color{purple}{(\star)}$ gives us:

\begin{align} 4y -3 &= 7\\ 4y & = 10\\ y &= \dfrac{10}{4}\\ &=\dfrac{5}{2} \end{align}

Therefore, the solution to the pair of original equations is $x=\dfrac{3}{2}, y = \dfrac{5}{2}$ or the point $\left(\dfrac{3}{2},\dfrac{5}{2}\right)$.

For case B, you must add the two equations together:

\begin{align} &20y -10x = 35\\ &2y +10x = 20\\ &\overline{22y\quad\;\,=55} \end{align}

We can now solve $22y=55$ for $y$ by dividing both sides by $22$. This gives $y= \dfrac{55}{22}=\dfrac{5}{2}$. Substituting this value of $y$ into either of $\color{purple}{(\dagger)}$ or $\color{purple}{(\star)}$ will enable us to find a solution for $x$. This time we shall use $\color{purple}{(\dagger \dagger)}$, to give us,

\begin{align} 5 +10x &= 20. &\color{purple}{(\dagger \dagger)}\\ 10x &= 15\\ x &= \dfrac{15}{10}\\ &= \dfrac{3}{2}.\\ \end{align}

The solution is then the point $\left(\dfrac{3}{2},\dfrac{5}{2}\right)$ or $x = \dfrac{3}{2}, y = \dfrac{5}{2}.$

Both cases produce the same solution. The intersection point is $\left(\dfrac{3}{2},\dfrac{5}{2}\right)$.

Here is a graphical solution for this example.

For this particular example, finding the point of intersection graphically is not too difficult. However, what if we wish to solve.

\begin{align} 17y - 3x &= 8\\ \text{and}\\ 9y +16x &= 13?\\ \end{align}

The graph looks like this:

As you can see, finding the intersection point graphically is not as straightforward as above and could result in a lack of precision. This demonstrates how it is important to know how to solve paired linear equations using the above method, rather than using graphs.