# Simultaneous Equations

Simultaneous Equation

This week will be talking about simultaneous equations. In simultaneous equations, we have linear simultaneous equations and the other that has one linear; one quadratic.

Simultaneous Linear Equations

Simultaneous linear equations are equations that contain two unknown quantities or variables which are required to be found.

Finding the two unknown quantities at the same time would be difficult without the methods of solving simultaneous equations.

Basically, we have two methods of solving simultaneously to find the value of the unknowns in linear equations.

1. Substitution method
2. Elimination method

In the substitution method, one of the unknown quantities is made the subject of formular in one of the equations; then it is substituted for in the other equation.

Example 1

Find the values of the unknowns in these equations:

x– 2y = 4

2x + 5y =17

Solution

Let the first equation be equation 1, and the second be equation 2.

x- 2y = 4——–1

2x + 5y =17—-2

(culled from Comprehensive Mathematics)

Now, we are going to make one of the unknowns the subject of formular in one of the equations.

In general, any of the unknowns can be made the subject of formular; you can make x the subject of formular in equation 1, you can also make y the subject of formular in equation 2. However there are some things you have to consider when choosing the unknown for the subject of formular.

Look out for the unknown whose coefficient is 1 in any of the two equations.

Do you know what I mean by coefficient?

In equation 1, 2 is the coefficient of y, and 1 is the coefficient of x.

In equation 2, 2 is the coefficient of x, and 5 is the coefficient of y.

So we have seen that in equation 1, x has coefficient of 1; so we are going to choose x.

If you make x the subject of formular in equation 1, we would have:

x = 4 + 2y

Now, we are going to substitute the value of x in the second equation.

So we have:

2(4 + 2y) + 5y = 17 (understood?)

Opening the bracket, we have:

8 + 4y + 5y = 17

This would give us:

8 + 9y = 17

Take 8 to the right side of the equation, you would have:

9y = 17 – 8

This would give us:

9y = 9

Now, divide both sides by 9, you would have:

y= 9/9

y = 1

Now that we know the value of y, we are going to find the value of x either of the equations.

From equation 1, x – 2y =4

Substituting for the value of y in the equation, we have:

x – 2(1) = 4

x – 2 = 4

Take  -2 over to the right side of the equation, you would have:

X = 4+2

X = 6.

To do double-check, you can also use the second equation to find the value of x.

2x + 5y = 17

2x + 5(1) = 17

2x + 5 = 17

2x = 17- 5

2x = 12

x= 12/2

x = 6.

So you see that either way, the value of x is the same.

Example 2

Find the values of the unknowns in these equations by solving simultaneously.

4x – 3y = 11

2x + 5y = 25

Solution

Name the equations accordingly.

4x – 3y = 11——1

2x + 5y = 25——2

Let’s make y the subject of the formular in equation 1

4x – 3y = 11——1

4x – 11 = 3y

y = 4x -11  ——–3

3

Now, we are going to substitute for y in equation 2.

2x + 5y = 25

2x + 5 (4x -11)  = 25

3

2x + 20x -55     = 25

3

3(2x) + 20x -55 =3(25)

6x + 20x – 55 = 75

26x = 75+55

26x = 130

x = 130/26

x = 5

Now we are going to find the value of y since we have found x. We can use any of the equations: 1, 2 or 3.

Let’s use equation 2.

2x + 5y = 25

2(5) + 5y = 25

10 + 5y  = 25

5y = 25-10

5y = 15

y = 15/5

y  = 3

Exercise

Solve the following equations by substitution

1.     3x +7y = 26

2x + 4y = 16

1.     x + 3y = 8

5x + 7y = 24

1.    y-5x = 12

y +4x =30