# Simultaneous equations 1

Last updated: September 28, 2021

## What is a simultaneous equation?

A simultaneous equation is where two algebraic expressions (typically in terms of $$x$$ and $$y$$) intersect with each other. When you solve for a simultaneous equation, you are solving for both $$x$$ and $$y$$. And they are co-ordinates.

Typically, a simultaneous equation looks like this:

Simultaneous equations can be solved in two ways:

• Substitution

• Elimination

## E2.5A: Derive and solve simultaneous equations in two unknowns

Simultaneous equations can be solved with the help of the substitution method. This method can be used to find the solution of a linear system of equations. We express one variable in terms of another, using one pair of equations and substituting that expression in the second equation.

### Worked examples of a simultaneous equation

Example 1: Solve the following pair of equations:

$$x+2y=6$$ and $$x-y=3$$

Step 1: Rearrange the first equation to express $$x$$ in terms of $$y$$.

The given equation is $$x+2y=6$$.

Rewrite the equation as $$x=6-2y$$.

Step 2: Substitute $$x=6-2y$$ in the second equation.

It is given that $$x-y=3$$.

It can be written as $$(6-2y)-y=3$$.

Step 3: Solve the above equation.

$$6-2y-y=3$$

$$6-3y=3$$

$$3y=6-3$$

$$3y=3$$

$$y=1$$

Therefore, the value of $$y$$ is $$1$$.

Step 4: Substitute $$y=1$$ in the equation $$x=6-2y$$.

$$x=6-2(1)$$

$$x=4$$

Therefore, the value of $$x$$ is $$4$$.

Example 2: The sum of the weights of Anisha and Arjun is $$60\text{ pounds}$$ and the difference is $$2$$. Find the weights of Anisha and Arjun.

Step 1: Write a simultaneous equation.

Let the weight of Anisha and Arjun be $$x\text{ pounds}$$ and $$y\text{ pounds}$$ respectively.

Therefore, the simultaneous equations are:

$$x+y=60$$

$$x-y=2$$

Step 2: Rearrange the first equation to express $$x$$ in terms of $$y$$.

The first equation is $$x+y=60$$.

It can be written as $$x=60-y$$.

Step 3: Substitute the value of $$x=60-y$$ in the second equation.

The second equation is $$x-y=2$$.

It can be written as $$(60-y)-y=2$$.

Step 4: Solve the above equation.

$$60-y-y=2$$

$$60-2y=2$$

$$2y=60-2$$

$$2y=58$$

$$y=29$$

Therefore, the value of $$y$$ is $$29$$.

Step 5: Substitute $$y=-29$$ in the equation $$x=60-y$$.

$$x=60-29$$

$$x=31$$.

Therefore, the value of $$x$$ is $$31$$.

So, the weight of Anisha is $$31\text{ pounds}$$ and the weight of Arjun is $$29\text{ pounds}$$.

## E2.5B: Derive and solve simultaneous equations involving one linear and one quadratic

Simultaneous equations can be solved with the help of the substitution method. While solving simultaneous equations of linear equations and quadratic equations, there are usually two pairs of answers. Substitute $$y$$ in the quadratic equation to create an equation that can be factorised and solved.

### Worked examples of solving a simultaneous equation involving a quadratic polynomial

Example 1: Solve the simultaneous equations given below:

$$y=x+1$$ and $$x^{2}+y^{2}=13$$.

Step 1: Substitute $$y=x+1$$ in the second equation.

The given equation is $$x^{2}+y^{2}=13$$.

It can be written as $$x^{2}+(x+1)^{2}=13$$.

Step 2: Expand the brackets.

$$x^{2}+(x+1)^{2}=13$$

$$x^{2}+x^{2}+2x+1=13$$

$$2x^{2}+2x-12=0$$

$$x^{2}+x-6=0$$

Step 3: Factorise the equation $$x^{2}+x-6=0$$.

It can be written as $$(x-2)(x+3)=0$$.

For $$(x-2)=0$$,

$$x-2=0$$

$$x=2$$

For $$(x+3)=0$$,

$$x+3=0$$

$$x=-3$$

Therefore, the values of $$x$$ are $$2$$ and $$-3$$.

Step 4: Substitute both the values of $$x$$ in the first equation to find the values of $$y$$.

It is given that $$y=x+1$$.

When $$x=2$$,

$$y=2+1$$

$$y=3$$

When $$x=-3$$,

$$y=-3+1$$

$$y=-2$$

Therefore, the values of $$y$$ are $$3$$ and $$-2$$.

So, the solutions of the given equations are $$(2,3)$$ and $$(-3,-2)$$.

Example 2: Solve the following simultaneous equations:

$$2x+3y=5$$ and $$2y^{2}+xy=12$$

Step 1: Rearrange the first equation to express $$x$$ in terms of $$y$$.

It is given that $$2x+3y=5$$.

It can be written as:

$$2x=5-3y$$

$$x=\frac{5-3y}{2}$$

Step 2: Substitute $$x=\frac{5-3y}{2}$$ in the second equation.

It is given that $$2y^{2}+xy=12$$

It can be written as:

$$2y^{2}+\left(\frac{5-3y}{2} \right )y=12$$

$$2y^{2}+\left(\frac{5y-3y^{2}}{2} \right )=12$$

$$4y^{2}+5y-3y^{2}=24$$

$$y^{2}+5y-24=0$$

Step 3: Factorise the equation $$y^{2}+5y-24=0$$.

It can be written as $$(y-3)(y+8)=0$$.

For $$y-3=0$$,

$$y-3=0$$

$$y=3$$

For $$y+8=0$$,

$$y+8=0$$

$$y=-8$$

Therefore, the values of $$y$$ are $$3$$ and $$-8$$.

Step 4: Substitute both the values of $$y$$ in the first equation to find the values of $$y$$.

It is given that $$2x+3y=5$$.

When $$y=3$$,

$$2x+3(3)=5$$

$$2x+9=5$$

$$2x=5-9$$

$$2x=-4$$

$$x=-2$$

When $$y=-8$$,

$$2x+3(-8)=5$$

$$2x-24=5$$

$$2x=5+24$$

$$2x=29$$

$$x=\frac{29}{2}$$

Therefore, the values of $$x$$ are $$x=-2$$ and $$x=\frac{29}{2}$$.

So, the solutions of the given equations are $$(-2,3)$$ and $$(\frac{29}{2},-8)$$.