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Give a system of equation that is more easily solved by substitution and a 

system of equations that is more easily solved by elimination.

Jan 13th, 2015

Example 1

Solve the following system of equations

Step 1

It's always better to label your equations so that you know which equation you're working with. Since we have two equations, let's label them as 1 and 2.


The first step in actually solving the system of equations using substitution is to express one variable in terms of another.

Let's use equation(1) and express y in terms of x in equation(1):

can also be written as

Step 3

Now we have y in terms of x and we can substitute for y in the equation(2)


Step 4

So now we only have a one variable equation which we can solve using the techniques we learned in the section on solving one variable equations.

Step 5

Now that we have a value for x, we can substitute it into the equation that we have for y and this is what we refer to as back substitution.

substituting for x


Step 6

So now we have solved for x as 2 and y as 0. Therefore, our coordinate point is (2,0). We can prove that these are the true values of x and y by substituting them back into the original system of equations.

substituting for x and y



\\ \left\{\begin{matrix} 3y+2x=6\\ 5y-2x=10 \end{matrix}\right. \\

We can eliminate the x-variable by addition of the two equations.

\\ 3y+2x=6\\ \underline{+\: 5y-2x=10} \\=8y\: \: \: \: \; \; \; \; =16 \\\begin{matrix} \: \: \: y\: \: \: \: \: \; \; \; \; \; =2 \end{matrix} \\

The value of y can now be substituted into either of the original equations to find the value of x

\\ 3y+2x=6 \\3\cdot {\color{green} 2}+2x=6 \\6+2x=6 \\x=0 \\

The solution of the linear system is (0, 2).

To avoid errors make sure that all like terms and equal signs are in the same columns before beginning the elimination.

If you don't have equations where you can eliminate a variable by addition or subtraction you directly you can begin by multiplying one or both of the equations with a constant to obtain an equivalent linear system where you can eliminate one of the variables by addition or subtraction.


\\ \left\{\begin{matrix} 3x+y=9\\ 5x+4y=22 \end{matrix}\right. \\

Begin by multiplying the first equation by -4 so that the coefficients of y are opposites

\\ {\color{green} -4}\cdot \left (3x+y\right )=9\cdot {\color{green} -4} \\ 5x+4y=22 \\\\-12x-4y=-36 \\\underline{+5x+4y=22 } \\=-7x\: \: \: \: \: \: \: \: \: \: =-14 \\\begin{matrix} \: \:\; \:\: x\: \: \: \: \: \: \: \: \: \: \:=2 \end{matrix} \\

Substitute x in either of the original equations to get the value of y

\\ 3x+y=9 \\3\cdot {\color{green} 2}+y=9 \\6+y=9 \\y=3 \\

The solution of the linear system is (2, 3)

Jan 13th, 2015

Jan 13th, 2015
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