Solving a system of equations by substitution of values

simultaneous equations, algebra problem, system of equations

A system of equations requires as many equations as there are variables in the equations.

One way to solve a system is to substitute values successively in each equation to eliminate variables.

Example:

Solve for x and y

Equation (1)      2x+y=16
Equation (2)      3(y+x)=33

There are two variables and we do have two equations so the system should have a solution.

We can find the value of y in terms of x in equation 1 by subtracting 2x from both sides of Equation (1)..

y=16-2x

We may then substitute this value for y into Equation (2).  Equation (2) will then look like this:

3[(16-2x) + x]=33

Removing the parentheses by distributing the 3 we get:

48-6x+3x=33

Simplifying the expression Equation (2) now looks like this:

3x=15

This means that x=5.

We may then substitute the value 5 for x in either equation.  Substituting 5 for x in Equation 1 we get:

2(5) + y=16.

To solve for y, we subtract 10 from both sides of the equation, leaving us with:

y=6.  

We have solved for both x and y and we may then substitute both values into both equations to check our work.

simultaneous equations, algebra problem, system of equations