Linear Equations – Learning Outcomes Use graphic, numeric, algebraic, and mental strategies to: Solve equations of the form 𝑎𝑥+𝑏 𝑒𝑥+𝑓 ± 𝑐𝑥+𝑑 𝑞𝑥+𝑟 =𝑘 Solve simultaneous equations with two unknowns and interpret the results. Solve simultaneous equations with three unknowns and interpret the results.

Use Graphic Strategies pg 58-66 Use Graphic Strategies Graphs are used often in business and science to depict and analyse trends. Draw a table showing the value of 50+20𝑥 between 𝑥 =0 and 𝑥=6. 𝑥 is called the independent variable and 50+20𝑥 is the dependent variable. Use your answer from the table to draw a graph of 50 +20𝑥 between 𝑥=0 and 𝑥=6. Use your graph to find the value of 50+20𝑥 when 𝑥 =2.5. Use your graph to find the value of 𝑥 which gives 50 +20𝑥=140.

Use Graphic Strategies pg 58-66 Use Graphic Strategies The table below shows the height of a plant in centimetres over a period of 35 days. Draw a graph to represent the data in the table. Calculate the slope of the graph. What does the slope represent? What does the y-intercept represent? Write a formula to represent the height of the plant over time. Day Height 8 5 12 10 16 15 20 24 25 28 30 32 35 36

Use Numeric Strategies pg 58-66 Use Numeric Strategies Consider the expression 50+20𝑥 again. Use an analogy to represent the situation. e.g. A plumber charges a €50 callout fee and additionally charges €20 per hour (i.e. per 𝑥). e.g. Your savings account holds €50 and you decide to deposit an extra €20 per week. e.g. The spell “Purge the Wicked” deals an initial 50 damage to the target, then another 20 damage per second.

Use Numeric Strategies pg 58-66 Use Numeric Strategies What does the plumber charge for two hours of work? If the plumber charges €90, how long did they work? How much money will you have saved after four weeks? How many weeks would you need to put money in your account to have saved €150? If “Purge the Wicked” expires after 20 seconds, how much damage does it deal in total?

Use Algebraic Strategies pg 58-66 Use Algebraic Strategies Consider the expression 50+20𝑥. Evaluate the expression when 𝑥=3 Evaluate the expression when 𝑥=1.6 Find the value of 𝑥 when 50+20𝑥=140 Find the value of 𝑥 when 50+20𝑥=205

Use Mental Strategies Mental strategies are used to quickly solve problems in your head. Usually you combine with another method to double- check your answer, since examiners can’t verify these strategies. e.g. we calculated that the plumber charges €90 for two hours of work. If we are then asked how many hours would cost €110, that’s €20 more euro than before, so one more hour. e.g. if the developers change “Purge the Wicked” to tick (deal damage) every two seconds instead, then its ongoing damage per second is halved (i.e. 10 DPS).

Solve Simultaneous Equations (2 vars) An equation with one variable is solvable on its own using all the expression and equation techniques we’ve studied. An equation with two variables does not have one solution, but represents a set of answers. e.g. 3𝑥+𝑦=7 has many possible solutions. If 𝑥=0, 𝑦=7 If 𝑥=1, 𝑦=4 If 𝑥=2, 𝑦=1 etc.

Solve SEs (2 vars) Graphically As before, we can use these solutions to generate a graph, which turns out to be linear.

Solve SEs (2 vars) Graphically Equations with two variables do not have a single answer. It takes two equations with two variables to get one answer. Graphically, this is the point of intersection of the two lines. 3𝑥+𝑦=7 5𝑥−4𝑦=6

Solve SEs (2 vars) Graphically

Solve SEs (2 vars) Graphically Solve the following simultaneous equations by drawing a graph: 𝑥+𝑦=4; 𝑥−𝑦=10 𝑥+𝑦=6; 2𝑥−4𝑦=12 5𝑥−2𝑦=9; 3𝑥+𝑦=1

Solve SEs (2 vars) by Elimination Elimination consists of adding or subtracting one equation from another to eliminate one variable. Elimination can only happen if each equation has the same coefficient for one variable, which may require some preparation: 3𝑥+𝑦=7 [×4] 5𝑥−4𝑦=6 12𝑥+4𝑦=28 Equations do not have same coefficients, so multiply first equation by 4. Now equations have same coefficient of 𝑦, with different signs, so add the equations.

Solve SEs (2 vars) by Elimination 12𝑥+4𝑦=28 5𝑥−4𝑦=6 17𝑥=34 ⇒𝑥=2 3 2 +𝑦=7 ⇒6+𝑦=7 ⇒𝑦=1 Adding the equations yields a new equation which has only one variable, which can be solved. To get the other variable, substitute the now known variable into an original equation.

Solve SEs (2 vars) by Substitution Elimination does not work for all types of simultaneous equations (e.g. quadratics, circles). Substitution is harder, but always works. It involves rearranging one equation in terms of one variable, then substituting into the other equation. 3𝑥+𝑦=7⇒𝑦=7−3𝑥 5𝑥−4𝑦=6 5𝑥−4 7−3𝑥 =6 Rearrange equation 1 in terms of 𝑦. Substitute this expression for 𝑦 in equation 2.

Solve SEs (2 vars) by Substitution 5𝑥−4 7−3𝑥 =6 ⇒5𝑥−28+12𝑥=6 ⇒17𝑥=34 ⇒𝑥=2 3 2 +𝑦=7 ⇒6+𝑦=7 ⇒𝑦=1 Substituting yields an equation in one variable that can be solved simply. Substitute this into an original equation to find the other variable.

Solve SEs (2 vars) by Algebra Solve the following simultaneous equations using either elimination or substitution: 𝑥+𝑦=4; 𝑥−𝑦=10 𝑥+𝑦=6; 2𝑥−4𝑦=12 5𝑥−2𝑦=9; 3𝑥+𝑦=1 5 𝑎 − 2 𝑏 =9; 3 𝑎 + 1 𝑏 =1 Why are the equations 2𝑥−𝑦=5 and 2𝑥−𝑦=3 not solvable simultaneously?

Solve SEs (3 vars) by Algebra One variable equations give single points. Two variable equations give lines. Three variable equations give planes. Just as lines don’t have single solutions, they represent a set of points on that line, planes represent a set of points on that plane.

Solve SEs (3 vars) by Algebra Intersecting two planes yields a line. Thus, two three-variable equations do not have a single solution. What if we add a third plane?

Solve SEs (3 vars) by Algebra The intersection of any pair of planes yields a line. The intersection of these lines yields a single point of intersection. Thus, three three-variable equations are required to get a single solution.

Solve SEs (3 vars) by Algebra 3𝑥−2𝑦+𝑧=9 [×2] 𝑥+4𝑦−5𝑧=−17 2𝑥+3𝑦+𝑧=−2 6𝑥−4𝑦+2𝑧=18 7𝑥−3𝑧=1 Solving these two at a time yields a representation of their line of intersection.

Solve SEs (3 vars) by Algebra 3𝑥−2𝑦+𝑧=9 [×3] 𝑥+4𝑦−5𝑧=−17 2𝑥+3𝑦+𝑧=−2 [×2] 9𝑥−6𝑦+3𝑧=27 4𝑥+6𝑦+2𝑧=−4 13𝑥+5𝑧=23 To get a final answer, we need to intersect two different planes, making sure to eliminate the same variable as before.

Solve SEs (3 vars) by Algebra 7𝑥−3𝑧=1 [×5] 13𝑥+5𝑧=23 [×3] 35𝑥−15𝑧=5 39𝑥+15𝑧=69 74𝑥=74 ⇒𝑥=1 The result is a pair of two- variable equations that can be solved simultaneously.

Solve SEs (3 vars) by Algebra 𝑥=1 ⇒7 1 −3𝑧=1 ⇒7−3𝑧=1 ⇒3𝑧=6 ⇒𝑧=2 3𝑥−2𝑦+𝑧=9 ⇒3 1 −2𝑦+2=9 ⇒3−2𝑦+2=9 ⇒𝑦=−2 This variable can be substituted into one of the two-variable equations. Both of these can be substituted into one of the plane equations.

Solve SEs (3 vars) by Algebra 2𝑥−5𝑦+𝑧=−5 3𝑥+5𝑦−2𝑧=7 5𝑥+5𝑦−5𝑧=0 7𝑥+5𝑦−3𝑧=16 3𝑥−5𝑦+2𝑧=−8 5𝑥+3𝑦−7𝑧=0

Solve SEs (3 vars) by Algebra In a triangle, angle B is 50 degrees less than 4 times angle A. Angle C is 40 degrees less than angle A. Find the measure of the three angles. In poker, chips are used as a substitute for currency. A white chip is worth €1, a red chip is worth €5, and a blue chip is worth €10. In a particular game, a Abby has 17 chips worth a total of €84. She has 6 more red chips than white chips. How many of each chip does she have?