Download Systems of Linear Equations and Their Solutions

Teacher – Mrs. Volynskaya
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
Exactly one ordered-pair solution
No solution
Infinitely many solutions
y
y
x
Exactly one solution
What This Means Graphically
The two lines intersect at one point.
The two lines are parallel.
The two lines are identical.
y
x
No Solution (parallel lines)
x
Infinitely many solutions
(lines coincide)
Example: Solving a System by Substitution
Solve by the substitution method: 5x – 4y = 9
x – 2y = -3
Solution Rewrite second equation as x =2y -3 and Sub. into first equation
This gives us an equation in one variable, 5(2y - 3) - 4y = 9.
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.
mor
Example: Solving a System by Substitution
Solve by the substitution method:
5x – 4y = 9
x – 2y = -3.
Solution
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).
Check. Take a moment to show that (5, 4) satisfies both given equations. The
solution set is {(5, 4)}.
Example: Solving a System by the Addition
Method
Solve by the addition method:
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.
mor
Solution
2x – 7y = -17
3x + 5y = 17
Steps 3 and 4
Multiply by 3.
Multiply by -2.
3•2x
-2•3x
– 3•7y = 3(-17)
+ (-2)5y = -2(17)
6x – 21y = -51
-6x + 10y = -34
Add the equations and solve for the remaining variable.
6x – 21y = -51
-6x – 10y = -34
-31y = -85
Add:
-31y = -85
-31
-31
y
= 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.
mor
Solution
2x – 7y = -17
3x + 5y = 17
Multiply by 5.
Multiply by 7.
5•2x – 5•7y = 5(-17)
7•3x + 7•5y = 7(17)
10x – 35y = -85
21x + 35y = 119
Add: 31x
x
= 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)}.
Examples
Determine the type of solution, then solve.
3x  2 y  1
1. 
x  y  3
4 x  6 y  12
2. 
6 x  9 y  36
4 x  5 y  3
3. 
2 x  3 y  7
2 x  y  3
4. 
4 x  2 y  5