MATH 31 U1A LESSON #4 SOLVING ABSOLUTE VALUE EQUATIONS NAME ANSWERS (π, π) (βπ, π) Practice 1. |2π₯ β 5| = 9 ππ β π = π x=7 β(ππ β π) = π βππ + π = π βππ = π x= β2 2. |1 β 3π₯| = 7 π β ππ = π x= β2 β(π β ππ) = π βπ + ππ = π ππ = π π x= π (βπ, π) π ( , π) π 3. |4π₯ β 1| + 5 = 8 |ππ β π| = π ππ β π = π ππ = π x= 1 β(ππ β π) = π βππ + π = π βππ = π π x = βπ U1A L4 ANS Abs Value Equations Page 1 of 4 π (β , π) π (π, π) MATH 31 U1A LESSON #4 SOLVING ABSOLUTE VALUE EQUATIONS NAME ANSWERS 4. 1 β |π₯ + 3| = β2 π=π+π π=π (π, βπ) (βπ, βπ) π = |π + π| π = β(π + π) π = βπ β π π = βπ 5. |2π₯ β 4| = 0 ππ β π = π ππ = π x= 2 (π, π) 6. |3π₯ β 2| + 5 = 3 |ππ β π| = βπ No Solution U1A L4 ANS Abs Value Equations Page 2 of 4 MATH 31 U1A LESSON #4 SOLVING ABSOLUTE VALUE EQUATIONS NAME ANSWERS Practice 7. |3π₯ + 5| = 2π₯ ππ + π = ππ π = βπ β(ππ + π) = ππ π = βπ CHECK |3(βπ) + 5| = 2(βπ) 10 β β10 CHECK |3(βπ) + 5| = 2(βπ) 2 β β2 There is NO solution 8. |1 β π₯| = 3π₯ β 2 π β π = ππ β π βππ = βπ π π= π CHECK π π |1 β | = 3 ( ) β 2 π π 1 1 = 4 4 π β π = β(ππ β π) = βππ + π ππ = π π π= π CHECK π π |1 β | = 3 ( ) β 2 π π 1 1 β β 2 2 π 1 ( , ) π 4 1 9. |π₯ β 2| = β 2 π₯ + 3 π πβπ=β π+π π π π=π π π ππ π = π× = π π CHECK ππ 1 ππ | β 2| = β ( ) + 3 π 2 π 4 10 18 =β + 3 6 6 4 4 = 3 3 U1A L4 ANS Abs Value Equations Page 3 of 4 π π π β π = β (β π + π) = π β π π π π π = βπ π π π = βπ× = βπ π CHECK (βπ, 4) 1 |βπ β 2| = β (βπ) + 3 2 4=4 ππ 4 ( , ) π 3 MATH 31 U1A LESSON #4 SOLVING ABSOLUTE VALUE EQUATIONS NAME ANSWERS 10. |π₯ + 6| = |2π₯| (π, 12) π + π = ππ π=π π + π = βππ ππ = βπ π = βπ (βπ, 4) 11. |1 β 3π₯| = |π₯ β 7| π β ππ = π β π βππ = βπ π=π π β ππ = βπ + π βππ = π π = βπ (βπ, 10) (π, 5) CHECK |1 β 3(π)| = |π β 7| CHECK |1 β 3(βπ)| = |βπ β 7| 5=5 10 = 10 U1A L4 ANS Abs Value Equations Page 4 of 4
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