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.
  2. Add the equations together
  3. Subtract one equation from another

Multiplying by a scaler

eqn 1 multiplied by 6 gives

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

Adding two equations

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

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