Simultaneous equations are those in which two or more equations must be solved at the same time.

Simultaneous equations can be solved using two methods:

1. Elimination method

2. Substitution method

**Elimination Method**

This entails eliminating one variable to solve for the other, and then inserting the answer into any of the original equations so that the other variable can be obtained.

It's worth noting, though, that only variables with the same co-efficient (s) can be eliminated.

**Example 1**

Solve for x and y in

$4x-2y=16$

$6x+2y=34$

**Solution:**

To eliminate y, we simply add the two equations

$\begin{matrix} 4x-2y=16\\ +(6x+2y=34)\\ \hline 10x+0=50 \end{matrix}$

$10x=50$

$x=5$

To obtain y, we insert the value of x in eqn 1

$4(5)-2y=16$

$20-2y=16$

$20-16=2y$

$2y=4$

$y=2$

__Example 2__

Solve for x and y in the equation

$30x+20y=100$

$20x+20y=20$

**Solution:**

To eliminate y, we subtract both eqn

$\begin{matrix} 30x+20y=100\\ -(20x+20y=20)\\ \hline 10x+0=80 \end{matrix}$

$10x=80$

$x=8$

Inserting the value of x in eqn 1

$30(8)+20y=100$

$240+20y=100$

$20y=100-240$

$20y=-140$

$y=-7$

Refer to this post for further information on the elimination method.

**Substitution Method**

This entails making a variable the subject of the formula of one of the equations and then inserting the derived expression for further simplification.

__Example 3__

By substitution, Solve for x and y in

$x+2y=13$

$2x-3y=5$

**Solution:**

First, we make x the subject of the formula in Eqn 1

$x=13-2y$

Inserting the value of x in Eqn 2

$2(13-2y)-3y=5$

$26-4y-3y=5$

$-7y=5-26$

$-7y=-21$

$\frac{-7y}{-7}=\frac{-21}{-7}$

$y=3$

Having obtained the value of y, we immediately insert it in Eqn 1 to get the value of x

$x+2(3)=13$

$x+6=13$

$x=13-6$

$x=7$

__Example 4__

Solve $23x+18y=51$

$9x+24y=3$

**Solution:**

Making x the subject of the formula in Eqn 2

$9x=3-24y$

$x=\frac{3-24y}{9}$

Now, let insert this in eqn 1

$23(\frac{3-24y}{9})+18y=51$

$\frac{69-552y}{9}+18y=51$

Taking the L.C.D at the L.H.S

$\frac{69-552y+162y}{9}=51$

$\frac{69-390y}{9}=51$

Cross multiplication

$69-390y=459$

$69-459=390y$

$-390=390y$

$y=-1$

Next, we insert the value of y in Eqn 2

$9x+24(-1)=3$

$9x-24=3$

$9x=3+24$

$9x=27$

$x=3$

Here are more examples for you to try out.

Having learned how to solve simultaneous equations, we turn our attention to Simultaneous linear and quadratic equations.

**Simultaneous Linear And Quadratic Equations**

These are simultaneous equations that consist of a linear equation and a quadratic equation which are meant to be solved simultaneously.

To solve this kind of equation, it is advisable to use the substitution method

__Example 5__

Solve for x and y in the equation:

$7x^2-28y^2=63$

$7x+14y=7$

**Solution:**

Let make x the subject of the formula in Eqn 2

$7x=7-14y$

$x=\frac{7-14y}{7}$

$x=\frac{7}{7}-\frac{14y}{7}$

$x=1-2y$

Inserting the value of x in eqn 1

$7(1-2y)^2-28y^2=63$

$7(1-4y+4y^2)-28y^2=63$

$7-28y+28y^2-28y^2=63$

$-28y=63-7$

$-28y=56$

$y=-2$

To obtain x, we replace $-2$ for y in eqn 2

$7x+14(-2)=7$

$7x-28=7$

$7x=7+28$

$7x=35$

$x=5$

__Example 6__

Solve for x and y in

$x^2-y^2=27$

$x+y=3$

**Solution:**

First and foremost, we make x the subject of the formula in Eqn 2

$x=3-y$

Now, let insert the expression of x in Eqn 1

$(3-y)^2-y^2=27$

$9-6y+y^2-y^2=27$

$-6y=27-9$

$-6y=18$

$y=-3$

$x+y=3$

$x-3=3$

$x=3+3$

$x=6$

Lastly, We looked at **word problem involving simultaneous equation**

__Example 7__

The sum of a woman and her daughter's age is 45 years. Five years ago, the woman was 6 times as old as his daughter. How old was the woman when she gave birth to the daughter?

**Solution:**

Let the woman age be x and let the daughter age be y.

From the first statement,

$$x+y=45$$...Eqn 1

From the second statement,

$x-5=6(y-5)$

$x-5=6y-30$

$x-6y=-30+5$

$$x-6y=-25$$.....Eqn 2

To determine the woman and her daughter age, we simply solve simultaneously

$x+y=45$

$x-6y=-25$

Now, let's subtract the eqns

$\begin{matrix} x+y=45\\ -(x-6y=-25)\\ \hline 0+7y=70 \end{matrix}$

$7y=70$

$y=10$

To obtain x, we replace 10 for y in eqn 1

#### $x+10=45$

#### $x=45-10$

$x=35$

To determine the age of the woman when she gave birth to her daughter, it is

$$35-10=25$$

Do you want to see more word problems with simultaneous equations? check this out.

__Related Posts__

- How to derive quadratic equations from quadratic roots
- Quadratic equations by completing the square
- Word problems involving quadratic equations

With that, we draw the curtain on the simultaneous equation. Did you find this post helpful? What other topics would you like us to assist you with? We'll be waiting for your feedback in the comments section.

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