The two main methods that you can solve simultaneous equations with two unknowns are:

● Algebraically

○ Substitution Method

○ Elimination Method

● Graphically

For Example we have the following set of linear simultaneous equations:

4x + 3y = 14 → Equation ①

5x + 7y = 11 → Equation ②

Let’s begin with solving this **algebraically** first. The first method I will be demonstrating is the **substitution method**.

Let’s pick x. You can also pick y. There is no restriction, but in this case, I'm choosing x.

Let’s choose Equation ①

4x + 3y = 14

4x + 3y = 14

● You begin with subtracting 3x from both sides.

4x + 3y - 3y = 14 - 3y

4x= 14 - 3y

● Now you divide both sides by 4.

4x ÷ 4 = (14 - 3y) ÷ 4

x= (14 - 3y) ÷ 4 → Equation ③

5[(14 - 3y) ÷ 4] + 7y = 11 →You substitute x with equation ③

[(70 - 15y) ÷ 4] + 7y = 11 → You multiply 5 with 14 - 3y

70 - 15y + 4(7y) = 11(4) → You multiply the entire equation by 4 to get rid of the fraction

70 - 15y + 28y = 44 → Subtract 70 from both sides

28y - 15y = 44 -70 → Collect the like terms on each side

13y = ⁻ 26 → Divide both sides by 13

y = ⁻ 2

x = (14 - 3y) ÷ 4

= (14 - 3[-2]) ÷ 4

= ( 14 + 6) ÷ 4

= 20 ÷ 4

= 5

We’ll be using the same set of linear simultaneous equations for this method as well. In contrast to the previous method, in this one, the main focus is to eliminate one variable, so that equation just contains one unknown variable and not two.

4x + 3y = 14 → Equation ①

5x + 7y = 11 → Equation ②

I’ll be choosing variable y this time.

If you look at both the equations, in Equation ①, the coefficient of y is 3 and in Equation ② it is 7. Hence we need to make sure that they are the same. In order to achieve that, we will multiply Equation ① by 7 and Equation ② by 3.

7×(4x + 3y = 14)

3×(5x + 7y = 11)

28x + 21y = 98

15x + 21y = 33

Since both the 21y’s in the equations have a positive sign, we subtract the two equations to completely eliminate y from the equation. If the 21y’s had opposite signs, then we’d be adding the two equations.

28x + 21y = 98

-(15x + 21y = 33)

28x - 15x = 98 - 33

13x = 65

x= 65/ 13

x= 5

4x + 3y = 14

4(5) + 3y = 14

20 + 3y = 14

20- 20 + 3y = 14 - 20

3y = -6

y= -6/3

y= -2

As you can see, both methods give the same answers for x and y.

When it comes to solving equations graphically, it focuses on plotting the equations on a grid and the intersection of the two equations basically gives the value of x and y.

4x + 3y = 14 → Equation ①

X- intercept is that point where the graph cuts the x-axis. For the x-intercept we put y = 0 in the equation.

4x + 3(0) = 14

4x = 14

x = 14/4

x = 3.5

Y-intercept is that point where your graph cuts the y-axis. For the y-intercept, we put x = 0 in the equation.

4(0) + 3y = 14

3y = 14

y = 14/3

y = 4.67

C( 3.5, 0) and D( 0, 4.67)

We repeat the same steps for equation 2

5x + 7y = 11 → Equation ②

For x-intercept:

5x + 7(0) = 11

5x = 11

x = 11/5

x = 2.5

For y-intercept:

5(0) + 7y = 11

7y = 11

y = 11/7

y = 1.57

A( 2.5, 0) and B( 0, 1.57)

Now that we have found the coordinates, we plot these on the axis and construct a graph for both the lines.

The intersection of the two lines gives us the solution for these simultaneous equations, which is the same as the answers found through the previous methods, x=5 and y=-2.