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# Systems of Linear Equations in Two Variables

Systems of Linear Equations and
Their Solutions

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 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).

Text Example

Determine whether (4, -1) is a solution of the system

x + 2y = 2
x – 2y = 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. 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)}.

Solving Linear Systems by Substitution

• 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.

Text Example

Solve by the substitution method:

5x – 4y = 9
x – 2y = -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 - 2y = -3 This is the second equation in the given system.
x = 2y - 3 Solve for x by adding 2y to both sides.

Step 2 Substitute the expression from step 1 into the other equation. We
substitute 2y - 3 for x in the first equation. Solve by the substitution method:

5x – 4y = 9
x – 2y = -3.

Solution
This gives us an equation in one variable, namely
5(2y - 3) - 4y = 9.
The variable x has been eliminated.

Step 3 Solve the resulting equation containing one variable.

 5(2y – 3) – 4y = 9 This is the equation containing one variable. 10y – 15 – 4y = 9 Apply the distributive property . 6y – 15 = 9 Combine like terms. 6y = 24 Add 15 to both sides. y = 4 Divide both sides by 6.

Solve by the substitution method:

5x – 4y = 9
x – 2y = -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 = 2y – 3.

 x = 2y – 3 Use the equation obtained in step 1. x = 2 (4) – 3 Substitute 4 for y. x = 8 – 3 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 Ax +
By = 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 ycoefficients
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.

Text Example

2x = 7y - 17
5y = 17 - 3x.

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

2x - 7y = -17
3x + 5y = 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 Steps 3 and 4 Add the equations and solve for the remaining variable. 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
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 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

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

Example Solution:

2 (2x + 3y = 4) multiply the first equation by 2
-4x - 6y = -1

4x + 6y = 8
-4x - 6y = -1

0 = 7 Add the two equations
No solution

Systems of Linear
Equations in Two
Variables

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