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One of the logical things about mathematics is that when you have one equation with one unknown
variable, you can always solve for the variable: 2x + 5 = 11 à 2x = 6 à x = 3. Similarly, when you
have two equations with two unknown variables, or three equations with three unknown variables, you can
always solve for the variables. In fact, a rule of algebra states that given x number of equations with x
number of variables, you can always solve for the variable(s).
Whenever two or more equations are presented, the group is known as a “system of equations.” There are
two methods to solve systems of equations; the one discussed here is called the substitution method. This
particular method teaches that after one of the equations has been solved for one of the variables in terms of
the other variable(s), you can substitute into the other equation(s) until all of the variables are found. Given
the system
2 x + 3 y = 5 Eq 1
x - 5 y = 9 Eq 2
scan the two equations for the variable that it would be easiest to isolate. In this case, the x in Eq 2 is easiest
because it has a coefficient of 1. So solve Eq 2 for x:
x – 5y = 9
x = 9 + 5y
Eq 2
Add 5y to each side
You can now represent x as 9 + 5y because the two are equivalent. Plug this new expression for x into the
other equation.The resulting equation is easy to solve because it looks like all of the other algebra
equations you’ve solved; it has only one variable!
2x + 3y = 5
2(9+ 5y) + 3y = 5
18 + 10y + 3y = 5
18 + 13y = 5
13y = -13
y = -1
Eq 1
9 + 5y is substituted for x into Eq 1.
Distribute the 2 throughout parentheses.
Combine like terms.
Subtract 18 from both sides to begin isolating the variable.
Divide both sides by 13 so that y is isolated.
You are half way there: y = -1. To find x, simply take the newly found value of y and plug it into one of
the original equations with both an x and a y.
2x + 3y = 5
2x + 3(-1) = 5
2x –3 = 5
2x = 8
x=4
Eq 1 randomly selected
Substitute -1 in for y.
Multiply.
Add 3 to each side.
Divide both sides by 2 to solve for x.
The answer can be written as point (x, y) = (4, -1) or by listing the variables separately, x = 4 and y = -1.
Note that this answer is not just any point. If the two original equations were graphed on a coordinate plane,
the two lines would cross at the point (4, -1). In this tutorial, the functions graphed will always be linear, so
there will be at most one point of intersection. However, if you graphed something other than two lines,
such as a circle and a line, there could be more than one point of intersection. So the answer you get when
you solve a system of equations is the point(s) at which the lines (or other non-linear functions) intersect.
Example 1: Solve this system of equations for x and y:
Eq 1:
Eq 2:
3x + 6y = 15
4x – 8y = 12
Solution:
We have chosen to solve Eq 1 for x because 3 divides evenly into both 6 and 15:
3x + 6y = 15
3x = 15 - 6y
15 - 6 y
x=
3
Eq 1
Subtract 6y from each side to begin isolating x
Divide each side by 3
x = 5 – 2y
Simplify fraction on right side
Now plug this new value of x into the other equation, Eq 2.
4x – 8y = 12
Eq 2
4( 5 - 2 y ) – 8y = 12
Plug 5 – 2y into the equation for x.
20 – 8y – 8y = 12
Distribute 4 throughout parentheses.
20 – 16y = 12
Combine like terms.
-16 y = -8
Subtract 20 from each side to begin isolating y.
y=
1
2
Divide both sides by -16.
Finally, take this value of y and plug it into one of the original equations to solve for x.
4x – 8y = 12
Eq 2
4 x - 8( 12 ) = 12
Plug ½ into the randomly selected equation for y.
4x – 4 = 12
Simplify.
4x = 16
Begin isolating x by adding 4 to each side.
x=4
Divide each side by 4.
Answer: The solution of this system of equations, the point at which the two lines cross, is
(4, ½) or x = 4, y = ½.
Given a system of equations with three unknowns and three equations, the substitution method still works
but has a few additional steps.
Example 2: Find the point (x, y, z) where the 3 lines intersect:
Eq 1: x – 3y + 6z = 21
Eq 2: 3x + 2y – 5z = -30
Eq 3: 2x – 5y + 2z = -6
Solution:
Solve one of the equations for one of the variables and plug this new value into the other two
equations. Then simplify the other two equations (note that these equations will still be in
terms of two variables). For this system, it will be easiest to solve Eq 1 for x because the
coefficient of the x is 1.
Solve Eq 1 for x: x – 3y + 6z = 21
x = 21 + 3y – 6z
x = 3y – 6z + 21
Isolate variable.
Rearrange terms alphabetically to be consistent
Plug this new value of x into Eq 2 and simplify:
3x + 2y – 5z = -30
3(3y – 6z + 21) + 2y – 5z = -30
9y – 18z + 63 + 2y – 5z = -30
11y – 23z = -93
Eq 2
Substitute for x.
Distribute the 3.
Combine terms.
Plug this new value of x into Eq 3 and simplify:
2x – 5y + 2z = -6
2(3y – 6z + 21) – 5y + 2z = -6
Eq 3
Substitute for x
6y – 12z + 42 – 5y + 2z = -6
y – 10z = - 48
Distribute the 2.
Combine terms.
The two equations created can be treated as a separate system of equations—there are two
equations and they have two variables! They can be solved for point (y, z) just as (x, y) was
found in Example 1.
New Eq 2: 11y – 23z = - 93
New Eq 3: y – 10z = - 48
Solve New Eq 3 for y because y has a coefficient of 1:
y – 10z = - 48
y = 10z - 48
New Eq 3
Add 10z to each side to isolate y.
Substitute this value of y in the other equation (New Eq 2) and solve for the variable z:
11y – 23z = -93
11(10z – 48) – 23z = -93
110z – 528 – 23z = -93
87z = 435
z=5
New Eq 2
Substitute (10z – 48) in for y.
Distribute 11 throughout parentheses.
Combine like terms.
Solve for z.
Now substitute z = 5 into one of the equations that has only two variables (New Eq 2 or New
Eq 3) and solve for y:
New Eq 3: y – 10z = - 48
y – 10(5) = - 48
y – 50 = - 48
y= 2
New Eq 3 is chosen because y has a
coefficient of 1.
Substitute 5 in for z.
Multiply.
Solve for y by adding 50 to each side.
Finally, substitute these values of y and z into one of the original equations and solve for x:
x – 3y + 6z = 21
x – 3(2) + 6(5) = 21
x – 6 + 30 = 21
x + 24 = 21
x = -3
Eq 1 chosen because coefficient of x is one
plug values of y and z into equation.
Multiply.
Add - 6 and 30.
Solve for x by subtracting 24 from each side.
Answer: The solution of this system of equations, the point at which the three lines cross,
is (-3, 2, 5) or x = -3, y = 2, and z = 5.