Equations with Numbers and
Unknowns on Both Sides
Slideshow 21, Mathematics
Mr Richard SasakiRoom 307
Objectivesβ’ Review algebraic vocabulary from
last lessonβ’ Solve by forming the unknown on one
side of the equation and a value on the other side
β’ Recall how to expand brackets and use this to solve such equations
β’ Consider numbers of solutions
Vocabulary ReviewDo you remember this from last time?
Terms Co-efficient Operator
Constant
3π₯+ΒΏ4Please try to get familiar with this!The expression above is linear.This means all terms have unknowns to the power 0 or 1.
Unknowns on Both SidesLetβs review solving for unknowns on both sides from last lesson.ExampleSolve β5 π₯ β5 π₯β2 π₯=β6Γ·(β2) Γ·(β2)π₯=3We try to get the unknown amount on one side and the constant number on the other side, even if we end up with negative numbers.
Numbers and Unknowns on Both SidesThe principle for equations with numbers and unknowns on both sides is the same. We separate the equation so numbers are on one side and algebraic terms are on the other side.ExampleSolve .We have 4 options of subtraction, 2 of which are more sensible. What are the two best things we could choose to do?It is best to either or . Why?
β2 β3 π₯
Numbers and Unknowns on Both SidesExampleSolve .If we choose to or , we will avoid negative numbers. Of course other options are okay too.5 π₯+2=3 π₯+8β2 β25 π₯=3 π₯+6β3 π₯ β3 π₯2 π₯=6Γ·2 Γ·2π₯=3
5 π₯+2=3 π₯+8β3 π₯ β3 π₯2 π₯+2=8β2 β22 π₯=6Γ·2 Γ·2π₯=3
Any order is okay!
Answers - Easyπ₯=2 π₯=5 π₯=3
π₯=4 π₯=β4 π₯=16
π¦=1 π₯=β1 π₯=β5π₯=3 π₯=1 π₯=6
π₯=9 π¦=β 12 π₯=0
Answers - Hardπ₯=Β±2π₯=Β±1 π₯=Β± 12
π₯=Β±5π₯=Β±2 π₯=Β± 4π₯=Β±3π₯=0 π₯=Β± 4π₯=β2π₯=2π₯=β 75 ππ π₯=5
Expanding BracketsLetβs review how to expand brackets. For linear style brackets, we multiply the term on the outside with the terms on the inside.
7 (3 π₯+4)ΒΏ21 π₯+28We will use this concept to solve equations.ExampleSolve .6 π¦β8=20β8 π¦+8 π¦ +8 π¦14 π¦β8=20+8 +814 π¦=28Γ·14 Γ·14π¦=2
Expanding BracketsLetβs try a harder question.Solve .12π₯+21=(36 π₯+54)β2112π₯ + 21 = 36 π₯ + 54 β 2112π₯ + 21 = 36 π₯ + 33β12 π₯ β12 π₯21=24 π₯+33β33 β33β12=24 π₯Γ·(β24) Γ·(β24)π₯=β 12
Answersπ₯=4 π₯=5
π₯=2 π₯=19
π₯=6 π₯=4
π₯=β2 π₯=4
Number of SolutionsWhen we try to solve an equation, we may find that it has solutions or infinite solutions.π₯+3=π₯ solutions π₯+2π₯=3 π₯Infinite solutions
ExampleState the number of solutions has.
3 (π₯+2 )=3π₯β53 π₯+6=3 π₯β5
6=β5As both sides do not equate, there are solutions.
Answers
Infinite solutions solutions
Infinite solutions solutions
solution, solutions, or