Home > Equations > Simultaneous equations: Elimination method

## Simultaneous equations: Elimination method

Simultaneous equations are sytems of equations where there are a number of variables.

Our task is to find the values of the variables that satisfy all the equations.
For this method we will solve equations of the type

$\displaystyle ax+by=c\rightarrow 1$

$\displaystyle dx+ey=f\rightarrow 2$

With these equations we can do one of three things

1. Multiply by a scaler.
3. Subtract one equation from another

Multiplying by a scaler

eqn 1 multiplied by 6 gives

$\displaystyle 6ax+6by=6c\rightarrow 3$

eqn 1 + eqn 2 gives

$\displaystyle (a+d)x+(b+e)y=c+f\rightarrow 4$

Subtract one equation from another

eqn 1 – eqn 2 gives

$\displaystyle (a-d)x+(b-e)y=c-f\rightarrow 5$

### Type 1 Simultaneous equations

With this type of problem the y terms are the same size but different signs.

To solve these we:

1. Add the equations together (this will cancel out the y terms)
2. Solve the resulting equation to find the value of x
3. Substitute the x value into one of the original equations
4. Solve the resulting equation to find the value of y
5. Check by substituting x and y into the other equation

Examples

Solve

$\displaystyle 3x+4y=18\rightarrow 1$

$\displaystyle 7x-4y=2\rightarrow 2$

This is a type 1 problem because we have two 4y’s, but one is positive and the other is negative.

So eqn 1 + eqn 2 gives

$\displaystyle 10x=20$

$\displaystyle \implies x=2$

Substitute $x=2$ into eqn 1

gives $\displaystyle 6+4y=18$

$\implies 4y=12$

$\implies y=3$

Check in eqn 2

$14-12=2$ true

Solve

$\displaystyle 2x-5y=13\rightarrow 1$

$\displaystyle 3x+5y=7\rightarrow 2$

This is a type 1 problem because we have two 5y’s, but one is positive and the other is negative.

So eqn 1 + eqn 2 gives

$\displaystyle 5x=20$

$\displaystyle \implies x=4$

Substitute $x=4$ into eqn 2 (it is easier to deal with positives)

gives $\displaystyle 12+5y=7$

$\implies 5y=-5$

$\implies y=-1$

Check in eqn 1

$8--5=13$ true

### Type 2 Simultaneous equations

With this type of problem the y terms are the same size and sign.

To solve these we:

1. Subtract one equation from the other (this will cancel out the y terms)
2. Solve the resulting equation to find the value of x
3. Substitute the x value into one of the original equations
4. Solve the resulting equation to find the value of y
5. Check by substituting x and y into the other equation

Examples

Solve

$\displaystyle 4x+2y=20\rightarrow 1$

$\displaystyle 7x+2y=23\rightarrow 2$

This is a type 2 problem because we have two +2y’s.

So eqn 2 – eqn 1 gives

$\displaystyle 3x=3$

$\implies x=1$

Substitute $x=1$ into eqn 1

gives $\displaystyle 4+2y=20$

$\implies 2y=16$

$\implies y=8$

Check in eqn 2

$\displaystyle 7+16=23$ true

Solve

$\displaystyle 2x-4y=6\rightarrow 1$

$\displaystyle 5x-4y=3\rightarrow 2$

This is a type 2 problem because we have two -4y’s.

So eqn 2 – eqn 1 gives

$\displaystyle 3x=-3\rightarrow 1$

$\implies x=-1$

Substitute $x=-1$ into eqn 1

gives $\displaystyle -2-4y=6$

$\implies 4y=-8$

$\implies y=-2$

Check in eqn 2

$\displaystyle -5--8=3$ true

### Type 3 Simultaneous equations

With this type of problem the y terms are different.

To solve these we:

1. Multiply each equation by the other equations positive y coefficient
2. Solve the resulting type 1 or type 2 problem

Examples

Solve

$\displaystyle 3x+4y=-5\rightarrow 1$

$\displaystyle 5x+2y=1\rightarrow 2$

Multiply eqn 1 by 2 (that is the number of y’s in eqn 2)

$\displaystyle 6x+8y=-10\rightarrow 3$

Multiply eqn 2 by 4 (that is the number of y’s in eqn 1)

$\displaystyle 20x+8y=4\rightarrow 4$

Equations 3 and 4 now make a type 2 problem

Subtracting eqn 3 from eqn 4 gives

$\displaystyle 14x=14$

$\displaystyle \implies x=1$

Substitute $x=1$ into eqn 2

$\displaystyle \implies 5+2y=1$

$\displaystyle \implies 2y=-4$

$\displaystyle \implies y=-2$

Check in eqn 1

$\displaystyle 5-4=1$ true

Solve

$\displaystyle 4x-2y=7\rightarrow 1$

$\displaystyle x+3y=10\rightarrow 2$

Multiply eqn 1 by 3 (that is the number of y’s in eqn 2)

$\displaystyle 12x-6y=21\rightarrow 3$

Multiply eqn 2 by 2 (that is the number of y’s in eqn 1)

$\displaystyle 2x+6y=20\rightarrow 4$

Equations 3 and 4 now make a type 1 problem

Adding eqn 3 and eqn 4 gives

$\displaystyle 14x=41\rightarrow 4$

$\displaystyle x=\frac{41}{14}$

Substitute $x=\frac{41}{14}$ into eqn 2

$\displaystyle \implies \frac{41}{14}+3y=10$

$\displaystyle \implies 3y=\frac{99}{14}\implies y=\frac{33}{14}$

Check in eqn 1

$\displaystyle \frac{164}{14}-\frac{66}{14}=7$ true