example 1 solve a system graphically graph the linear system and estimate the solution. then check...
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EXAMPLE 1 Solve a system graphically
Graph the linear system and estimate the solution. Then check the solution algebraically.
4x + y = 8
2x – 3y = 18
Equation 1
Equation 2
SOLUTION
Begin by graphing both equations, as shown at the right. From the graph, the lines appear to intersect at (3, – 4). You can check this algebraically as follows.
EXAMPLE 1 Solve a system graphically
Equation 1 Equation 2
4x + y = 8
4(3) + (– 4) 8=?
=?12 – 4 8
8 = 8
2x – 3y = 18
=?2(3) – 3( – 4) 18
=?6 + 12 18
18 = 18
The solution is (3, – 4).
EXAMPLE 2 Solve a system with many solutions
Solve the system. Then classify the system as consistent and independent,consistent and dependent, or inconsistent.
4x – 3y = 8
8x – 6y = 16
Equation 1
Equation 2
SOLUTION
The graphs of the equations are the same line. So, each point on the line is a solution, and the system has infinitely many solutions. Therefore, the system is consistent and dependent.
EXAMPLE 3 Solve a system with no solution
Solve the system. Then classify the system as consistent and independent,consistent and dependent, or inconsistent.
2x + y = 4
2x + y = 1
Equation 1
Equation 2
SOLUTION
The graphs of the equations are two parallel lines.Because the two lines have no point of intersection, the system has no solution. Therefore, the system is inconsistent.
EXAMPLE 4 Standardized Test Practice
SOLUTION
Equation 1 (Option A)
y = 1 x + 30
Equation 2 (Option B)
EXAMPLE 4 Standardized Test Practice
y = x2.5
To solve the system, graph the equations y = x + 30 and y = 2.5x, as shown at the right.
EXAMPLE 4 Standardized Test Practice
Notice that you need to graph the equations only in the first quadrant because only nonnegative values of x and y make sense in this situation.
The lines appear to intersect at about the point (20, 50). You can check this algebraically as follows.
Equation 1 checks.
Equation 2 checks.
50 = 20 + 30
50 = 2.5(20)
ANSWERThe total costs are equal after 20 rides.
The correct answer is B.
EXAMPLE 1 Use the substitution method
Solve the system using the substitution method.
2x + 5y = – 5
x + 3y = 3
Equation 1
Equation 2
SOLUTION
STEP 1 Solve Equation 2 for x.
x = – 3y + 3 Revised Equation 2
EXAMPLE 1 Use the substitution method
STEP 2Substitute the expression for x into Equation 1 and solve for y.
2x +5y = – 5
2(– 3y + 3) + 5y = – 5
y = 11
Write Equation 1.
Substitute – 3y + 3 for x.
Solve for y.
STEP 3Substitute the value of y into revised Equation 2 and solve for x.
x = – 3y + 3
x = – 3(11) + 3
x = – 30
Write revised Equation 2.
Substitute 11 for y.
Simplify.
EXAMPLE 1 Use the substitution method
CHECK Check the solution by substituting into the original equations.
2(– 30) + 5(11) – 5=? Substitute for x and y. =?– 30 + 3(11) 3
Solution checks. 3 = 3– 5 = – 5
The solution is (– 30, 11).
ANSWER
EXAMPLE 2 Use the elimination method
Solve the system using the elimination method.
3x – 7y = 10
6x – 8y = 8
Equation 1
Equation 2
SOLUTION
Multiply Equation 1 by – 2 so that the coefficients of x differ only in sign.
STEP 1
3x – 7y = 10
6x – 8y = 8
– 6x + 14y = 20
6x – 8y = 8
EXAMPLE 2 Use the elimination method
STEP 2Add the revised equations and solve for y. 6y = – 12
y = – 2STEP 3Substitute the value of y into one of the original equations. Solve for x.
3x – 7y = 10
3x – 7( – 2) = 10
3x + 14 = 10
x = 43
– Solve for x.
Simplify.
Substitute – 2 for y.
Write Equation 1.
EXAMPLE 2 Use the elimination method
The solution is ( , – 2)43
–
ANSWER
CHECK
You can check the solution algebraically using the method shown in Example 1. You can also use a graphing calculator to check the solution.
EXAMPLE 3 Standardized Test Practice
SOLUTION
Write verbal models for this situation.
Equation 1
EXAMPLE 3 Standardized Test Practice
Equation 2
EXAMPLE 3 Standardized Test Practice
STEP 2 Write a system of equations.
Equation 1
Equation 2
5x + 7y = 2500
8x + 12y = 4200
Total cost for all T-shirts
Total revenue from all T-shirts sold
STEP 3Solve the system using the elimination method.
Multiply Equation 1 by – 8 and Equation 2 by 5 so that the coefficients of x differ only in sign.
5x + 7y = 2500
8x + 12y = 4200
– 40x – 56y = – 20,000
40x + 60y = 21,000
Add the revised equations and solve for y. 4y = 1000y = 250
EXAMPLE 3 Standardized Test Practice
Substitute the value of y into one of the original equations and solve for x.
5x + 7y = 2500
5x + 7(250) = 2500
5x + 1750 = 2500x = 150
Write Equation 1.
Substitute 250 for y.
Simplify.
Solve for x.
The school sold 150 short sleeve T-shirts and 250 long sleeve T-shirts.
ANSWERThe correct answer is C.
EXAMPLE 4 Solve linear systems with many or no solutions
Solve the linear system.
a. x – 2y = 43x – 6y = 8
b. 4x – 10y = 8– 14x + 35y = – 28
SOLUTION
a. Because the coefficient of x in the first equation is 1, use the substitution method.
Solve the first equation for x.
x – 2y = 4
x = 2y + 4
Write first equation.
Solve for x.
EXAMPLE 4 Solve linear systems with many or no solutions
Substitute the expression for x into the second equation.
3x – 6y = 8
3(2y + 4) – 6y = 8
12 = 8
Write second equation.
Substitute 2y + 4 for x.
Simplify.
Because the statement 12 = 8 is never true, there is no solution.
ANSWER
EXAMPLE 4 Solve linear systems with many or no solutions
b.Because no coefficient is 1 or – 1, use the elimination method.
Multiply the first equation by 7 and the second equation by 2.
4x – 10y = 8
– 14x + 35y = – 28
28x – 70y = 56
– 28x + 70y = – 56
Add the revised equations. 0 = 0
ANSWER
Because the equation 0 = 0 is always true, there are infinitely many solutions.