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Solving Simple Simultaneous Equations Slideshow 12, Mr Richard Sasaki Mathematics, Room 307 Objectives β’ Review how to subtract polynomials in vertical form β’ Look at how simultaneous equations work β’ Solve simultaneous equations with one coefficient and term the same. Subtracting Polynomials Review Subtracting Polynomials can be done vertically, in the same way as subtracting two numbers. We can calculate from left to right or right to left. Examples 19x + 8y 12π₯ + 9π¦ 7π₯ - π¦ Try the worksheet! 9π₯ β 3π¦ β3π₯ +12π₯5π¦ β 8π¦ Answers π₯ + 3π¦ 3π₯ + 5π¦ β5π₯ β 3π¦ β7π₯ + 3π¦ 3π¦ 7π₯ β 2π¦ 11π₯ β 3π¦ 3½π₯ + 5π¦ 3π₯ β 9π¦ β3π₯ + 3π¦ 5π₯ + 8π¦ β4π₯ + 5π¦ Simultaneous Equations Linear simultaneous equations with two unknowns usually have two only one possibility for each unknown. Each unknown only represents one number. You may have realised this from last lesson. Simultaneous means βat the same timeβ. These two (or more) equations work together at the same time. So how do we solve them? Simultaneous Equations Today, we will look at examples where a coefficient and term are the same in each equation. Example Solve the simultaneous equations below. What can we do to β π₯ + 5π¦ = 8 solve for x or y? β‘π₯ + 3π¦ = 6 Yes, letβs subtract the 2π¦ = 2 β οΌ top from the bottom! β‘ π¦ = 1 Then we can remove β π₯ + 5π¦ = 8 x and solve for y. Now we need to substitute y = 1 into π₯ + 5(1) = 8 β or β‘. Either are π₯ + 5 = 8 So π₯ = 3 and π¦ = 1. okay! Simultaneous Equations Letβs try another example. Example Solve the simultaneous equations below. Yes, letβs subtract the β 3π₯ + 5π¦ = 2 top from the bottom β‘3π₯ β 2π¦ = 16 again. (Subtracting the β οΌ β‘ 7π¦ = β14 π¦ = β2 Substitute π¦ = __ β 3π₯ + 5π¦ = 2 into whichever 3π₯ + 5(β2) = 2 looks easiest. 3π₯ β 10 = 2 3π₯ = 12 π₯ = 4 bottom from the top is also okay). Once again, we can remove π₯ and solve for π¦. So π₯ = 4 and π¦ = β2. Simultaneous Equations So for a summary, to solve simultaneous equations like this, we subtract one from another (doesnβt matter which) and solve for the other unknown. Then we substitute this into one of the equations (whichever you like) and find the result for the other. Remember to write your answer clearly at the end! Good luck and try the worksheets! Answers β Easy & Medium 2π¦ = 4 It disappears. 2 0 3 8 5 2 -3 1 5π₯ = 5 It disappears. 4π¦ = 16 4 3 π = 4, π = 1 2 3 -9 7 2 9 Answers β Hard 2 3 3 1 5 3 5 3 2 5 37 7 5 4 10 7