We have seen that all equations in the form Ax + By = C are straight lines when graphed. Two such equations, such as those listed above, are called a system of linear equations. A solution to a system of linear equations is an ordered pair that satisfies all equations in the system. For example, (3, 4) satisfies the system x + y = 7 x – y = -1
(3 + 4 is, indeed, 7.)
(3 – 4 is indeed, -1.)
Thus, (3, 4) satisfies both equations and is a solution of the system. The solution can be described by saying that x = 3 and y = 4. The solution can also be described using set notation. The solution set to the system is {(3, 4)} - that is, the set consisting of the ordered pair (3, 4).

Determine whether (4, -1) is a solution of the system x + 2 y = 2 x – 2 y = 6.
Solution Because 4 is the x -coordinate and -1 is the y -coordinate of (4, -1), we replace x by 4 and y by -1.
x + 2 y = 2
?
x – 2 y = 6
?
4 + 2(-1) = 2 4 – 2(-1) = 6
?
4 + (-2) = 2
?
4 – (-2) = 6
2 = 2 true
?
4 + 2 = 6
6 = 6 true
The pair (4, -1) satisfies both equations: It makes each equation true. Thus, the pair is a solution of the system. The solution set to the system is {(4, -1)}.

• Solve either of the equations for one variable in terms of the other. (If one of the equations is already in this form, you can skip this step.)
• Substitute the expression found in step 1 into the other equation. This will result in an equation in one variable.
• Solve the equation obtained in step 2.
• Back-substitute the value found in step 3 into the equation from step 1. Simplify and find the value of the remaining variable.
• Check the proposed solution in both of the system's given equations.

Solve by the substitution method:
5 x – 4 y = 9 x – 2 y = -3.
Solution
Step 1 Solve either of the equations for one variable in terms of the other.
We begin by isolating one of the variables in either of the equations. By solving for x in the second equation, which has a coefficient of 1, we can avoid fractions.
x - 2 y = -3 This is the second equation in the given system.
x = 2 y - 3 Solve for x by adding 2 y to both sides.
Step 2 Substitute the expression from step 1 into the other equation . We substitute 2 y - 3 for x in the first equation.
x = 2 y – 3 5 x – 4 y = 9

Solve by the substitution method:
5 x – 4 y = 9 x – 2 y = -3.
Solution
This gives us an equation in one variable, namely
5(2 y - 3) - 4 y = 9.
The variable x has been eliminated.
Step 3 Solve the resulting equation containing one variable.
5(2 y – 3) – 4 y = 9 This is the equation containing one variable.
10 y – 15 – 4 y = 9
6 y – 15 = 9
Apply the distributive property.
Combine like terms.
6 y = 24 y = 4
Add 15 to both sides.
Divide both sides by 6.

Solve by the substitution method:
5 x – 4 y = 9 x – 2 y = -3.
Solution
Step 4 Back-substitute the obtained value into the equation from step 1 .
Now that we have the y -coordinate of the solution, we back-substitute 4 for y in the equation x = 2 y – 3.
x = 2 y – 3
Use the equation obtained in step 1.
x = 2 (4) – 3 x = 8 – 3
Substitute 4 for y.
Multiply.
x = 5 Subtract.
With x = 5 and y = 4, the proposed solution is (5, 4).
Step 5 Check . Take a moment to show that (5, 4) satisfies both given equations. The solution set is {(5, 4)}.

• If necessary, rewrite both equations in the form A x +
B y = C.
• If necessary, multiply either equation or both equations by appropriate nonzero numbers so that the sum of the x -coefficients or the sum of the y coefficients is 0.
• Add the equations in step 2. The sum is an equation in one variable.
• Solve the equation from step 3.
• Back-substitute the value obtained in step 4 into either of the given equations and solve for the other variable.
• Check the solution in both of the original equations.

Solve by the addition method:
2 x = 7 y - 17
5 y = 17 - 3 x .
Solution
Step 1 Rewrite both equations in the form Ax + By = C.
We first arrange the system so that variable terms appear on the left and constants appear on the right. We obtain
2 x - 7 y = -17
3 x + 5 y = 17
Step 2 If necessary, multiply either equation or both equations by appropriate numbers so that the sum of the x-coefficients or the sum of the y-coefficients is 0.
We can eliminate x or y . Let's eliminate x by multiplying the first equation by 3 and the second equation by -2.

Solution
2 x
3 x
–
+
7 y
5 y
=
=
-17
17
Multiply by 3.
Multiply by -2.
3• 2 x
-2• 3 x
–
+
3• 7 y =
(-2) 5 y =
3 (-17)
-2 (17)
6 x
-6 x
–
–
21 y
10 y
=
=
-51
-34
Steps 3 and 4 Add the equations and solve for the remaining variable .
Add:
6 x
-6 x
–
–
21 y
10 y
-31 y
=
=
=
-51
-34
-85
-31 y
-31 y
=
=
-85
-31
85/31
Divide both sides by -31.
Simplify.
Step 5 Back-substitute and find the value for the other variable.
Backsubstitution of 85/31 for y into either of the given equations results in cumbersome arithmetic. Instead, let's use the addition method on the given system in the form Ax + By = C to find the value for x . Thus, we eliminate y by multiplying the first equation by 5 and the second equation by 7.

Solution
2 x
3 x
–
+
7 y
5 y
=
=
-17
17
Multiply by 5.
Multiply by 7.
5• 2 x
7• 3 x
–
+
5• 7 y
7• 5 y
=
=
5 (-17)
7 (17)
Add:
10 x
21 x
–
+
35 y
35 y
31 x x
=
=
=
=
-85
119
34
34/31
Step 6 Check.
For this system, a calculator is helpful in showing the solution (34/31, 85/31) satisfies both equations. Consequently, the solution set is {(34/31, 85/31)}.

The number of solutions to a system of two linear equations in two variables is given by one of the following.
Number of Solutions
Exactly one ordered-pair solution
No solution
Infinitely many solutions y y
What This Means Graphically
The two lines intersect at one point.
The two lines are parallel.
The two lines are identical.
y
Exactly one solution x x
No Solution (parallel lines) x
Infinitely many solutions
(lines coincide)

Solve the system 2x + 3y = 4
-4x - 6y = -1
Solution:
2 (2x + 3y = 4) multiply the first equation by 2
-4x - 6y = -1
4x + 6y = 8
-4x - 6y = -1
0 = 7
No solution
Add the two equations