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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 2
Substitute the expression for x into Equation 1 and solve
for y.
2x +5y = –5 Write Equation 1.
2(–3y + 3) + 5y = –5
y = 11
Substitute –3y + 3 for x.
Solve for y.
STEP 3
Substitute the value of y into revised Equation 2 and
solve for x.
x = –3y + 3
Write revised Equation 2.
x = –3(11) + 3 Substitute 11 for y.
x = –30
Simplify.
EXAMPLE 1
Use the substitution method
ANSWER
The solution is (– 30, 11).
CHECK Check the solution by substituting into the
original equations.
2(–30) + 5(11) =? –5
–5 = –5
Substitute for x and y. –30 + 3(11) =? 3
Solution checks.
3=3
EXAMPLE 2
Use the elimination method
Solve the system using the elimination method.
Equation 1
3x – 7y = 10
Equation 2
6x – 8y = 8
SOLUTION
STEP 1
Multiply Equation 1 by – 2 so that the coefficients of x
differ only in sign.
3x – 7y = 10
–6x + 14y = 220
6x – 8y = 8
6x – 8y = 8
EXAMPLE 2
Use the elimination method
STEP 2
Add the revised equations and solve for y. 6y = –12
y = –2
STEP 3
Substitute the value of y into one of the original
equations. Solve for x.
3x – 7y = 10
3x – 7(–2) = 10
3x + 14 = 10
4
x =– 3
Write Equation 1.
Substitute –2 for y.
Simplify.
Solve for x.
EXAMPLE 2
Use the elimination method
ANSWER
The solution is ( –
4
, –2)
3
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.
GUIDED PRACTICE
for Examples 1 and 2
Solve the system using the substitution or the
elimination method.
1. 4x + 3y = –2
x + 5y = –9
2. 3x + 3y = –15
5x – 9y = 3
ANSWER
ANSWER
The solution is (1,–2).
The solution is (– 3 , –2)
GUIDED PRACTICE
for Examples 1 and 2
Solve the system using the substitution or the
elimination method.
3. 3x – 6y = 9
–4x + 7y = –16
ANSWER
The solution is (11, 4)