Simultaneous equations - Intermediate & Higher tier – WJECSolving simultaneous equations graphically

Simultaneous equations require algebraic skills to find the value of unknowns within two or more equations that are true at the same time.

Part ofMathsAlgebra

Solving simultaneous equations graphically

Simultaneous equations can be solved algebraically or graphically. Knowledge of plotting linear and quadratic graphs is needed to solve equations graphically.

To find solutions from graphs, look for the point where the two graphs cross one another. This is the solution point. For example, the solution for the graphs \(y = x + 1\) and \(x + y = 3\) is the coordinate point (1, 2).

Two graphs on a grid. 1 is labelled y = x + 1, the other is labelled x + y = 3. The 2 graphs cross at the coordinate point 1, 2

The solution to these equations is \(x = 1\) and \(y = 2\).

Solving linear equations graphically

Example

Solve the simultaneous equations \(x + y = 5\) and \(y = x + 1\) using graphs.

To solve this question, first construct a set of axes, making sure there is enough room to plot the 2 graphs.

A sheet of blank graph paper

Now draw the graphs for \(x + y = 5\) and \(y = x + 1\).

To draw these graphs, use a table of values:

\(y = x + 1\)

\(x\)-10123
\(y\)01234
\(x\)
-1
0
1
2
3
\(y\)
0
1
2
3
4

\(x + y = 5\)

\(x\)-10123
\(y\)65432
\(x\)
-1
0
1
2
3
\(y\)
6
5
4
3
2

Plot these graphs onto the axes and label each graph.

Two graphs on a grid. 1 is labelled y = x + 1, the other is labelled x + y = 5. The 2 graphs cross at the point of intersection at 2, 3

The point of intersection is (2, 3) which means \(x = 2\) and \(y = 3\).

Solving linear and quadratic equations graphically - Higher

Simultaneous equations that contain a quadratic and equation can also be solved graphically. As with solving algebraically, there will usually be two pairs of solutions.

Example

Solve the simultaneous equations \(y = x^2\) and \(y = x + 2\).

\(y = x^2\)

\(x\)-3-2-10123
\(y\)9410149
\(x\)
-3
-2
-1
0
1
2
3
\(y\)
9
4
1
0
1
4
9

\(y = x + 2\)

\(x\)-10123
\(y\)12345
\(x\)
-1
0
1
2
3
\(y\)
1
2
3
4
5

Plot the graphs on the axes and look for the points of intersection.

Two graphs on a grid. 1 is labelled y = x + 2, the other is labelled y = x squared. The two points of intersection are at 2, 4 and -1, 1

The two points of intersection are at (2, 4) and (-1, 1) so \(x = 2\) and \(y = 4\), and \(x = -1\) and \(y = 1\).