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Simultaneous Equations, Solving by Substitution,
Solving by Elimination


The  solving simultaneous equations introduction  page explained how we can solve 2 simultaneous linear equations with 2 variables together. and that a combined solution will graphically be a point of intersection of the 2 straight lines that the equations represent.


This page will show the process of solving two simultaneous equations, with the methods of solving by substitution and solving by elimination.

Both are common methods used for solving when dealing with simultaneous equations in Math.






Solving by Substitution


When dealing with 2 equations involving 2 variables, such as.

x + 2y = 4
3x \space {\text{–}} \space y = 5


When solving by substitution we want to put one of the equations into the form  “x =”  or  “y =“.

Then ‘substitute’ said equation into the other equation in place of the relevant variable, then solve to obtain a value for the other variable.
This value obtained can then be used in the other equation to solve fully for both variables in the equations.


So if we follow this approach with the equations above, we can start by labelling them with numbers.

x + 2y = 4       [1]
3x \space {\text{–}} \space y = 5        [2]


Subtracting  2y  from both sides of  [1]  will give us an equation in the form  “x =“.

x + 2y \space {\text{–}} \space 2y = 4 \space {\text{–}} \space 2y     =>     x = 4 \space {\text{–}} \space 2y


Now substituting this into  [2]  we have.

3(4 \space {\text{–}} \space 2y) \space {\text{–}} \space y = 5    =    12 \space {\text{–}} \space 6y \space {\text{–}} \space y = 5

12 \space {\text{–}} \space 7y = 5    ,    {\text{-}}7y = {\text{-}}7     =>     y = 1

Now substituting this  y  value into  [1]  can give us  x.

x + 2(1) = 4     =>     x + 2 = 4    ,    x = 2




Example    


(1.1) 

Solve the simultaneous equations:

\space x + y = 6
2x \space {\text{–}} \space 3y = 7

by solving by substitution.


Solution   

If we begin by labelling each equation.

\space x + y = 6         [1]
2x \space {\text{–}} \space 3y = 7       [2]

We can see that subtracting  y  from both sides of  [1]  will get the equation in terms of just  x.

x + y \space {\text{–}} \space y = 6 \space {\text{–}} \space y     =>     x = 6 \space {\text{–}} \space y


Substituting this into  [2]  gives.

2(6 \space {\text{–}} \space y) \space {\text{–}} \space 3y = 7    =    12 \space {\text{–}} \space 2y \space {\text{–}} \space 3y = 7

12 \space {\text{–}} \space 5y = 7    ,    {\text{-}}5y = {\text{-}}5     =>     y = 1

Substituting this  y  value into  [1]  to obtain  x.

x + 1 = 6    ,    x = 5

It’s generally good practice to also check that the values for the variables also work for the other equation, to be sure they are correct.

So here we can input the values in to equation  [2].


2(5) \space {\text{–}} \space 3(1) = 7    =>    10 \space {\text{–}} \space 3 = 7       ( CORRECT )





Solving by Elimination


The idea behind solving by elimination is similar to solving by substitution.

We again begin by looking to find an initial value for one of the variables, by adding or subtracting one equation from the other, or where appropriate a multiple of one equation.


If we consider solving the simultaneous equations,

5x + y = 7
x + y = 3

Subtracting the 2nd equation from the 1st eliminates the  y  variable.

\begin{array}{r} \space{5x + y = 7}\space\space\\ {\text{–}}\space\space\space\space{x + y = 3}\space\space\\ \hline {4x \space\space{\color{white}+ y} = 4}\space\space \end{array}

We can see from the subtraction that  x = 1.

Now using the 2nd equation.  1 + y = 3
So  y = 2.

Checking in the 1st equation.

5(1) + 2 = 7    =>    5 + 2 = 7       ( CORRECT )




Example    


(2.1) 

Solve the simultaneous equations:

4x + y = 8
x \space {\text{–}} \space 2y = 11

by elimination.


Solution   

Adding or subtracting the equations as they sit, won’t eliminate a variable in one of the equations.

But multiplying the 1st equation by  2,  then adding the equations together will achieve a variable elimination.

4x + y = 8   ( × 2 )     =     8x + 2y = 16

\begin{array}{r} \space\space{8x + 2y = 16}\space\space\\ +\space\space\space{x \space {\text{–}} \space 2y = 11}\space\space\\ \hline {9x \space\space{\color{white}+ 2y} = 27}\space\space \end{array}


The result tells us that  x = 3.
Which we can input into the 2nd equation to obtain  y.

3 \space {\text{–}} \space 2y = 11

3 \space {\text{–}} \space 3 \space {\text{–}} \space 2y = 11 \space {\text{–}} \space 3     =>     {\text{-}}2y = 8     =>     y = {\text{-}}4

Checking with 1st equation:

4(3) + ({\text{-}}4) = 8     =>     12 \space {\text{–}} \space 4 = 8       ( CORRECT )





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