# Simultaneous equations 1

**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 $$(\frac{29}{2},-8)$$.