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The intersection of two lines. Back in term 1, exercise 3.10, we looked at linear simultaneous equations. We noted there are 3 main ways to solve simultaneous equations: Elimination (what we will mainly use) Substitution (if one or both equations start with y = )
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The intersection of two lines. • Back in term 1, exercise 3.10, we looked at linear simultaneous equations. • We noted there are 3 main ways to solve simultaneous equations: • Elimination (what we will mainly use) • Substitution (if one or both equations start with y = ) • Graphically (remember everything must be drawn to scale)
Example 1 Find the point of intersection of the lines 3x – 4y + 22 = 0 and 5x + 6y – 14 = 0. 3x – 4y + 22 = 0… 5x + 6y – 14 = 0… × 3 9x – 12y + 66 = 0… × 2 10x + 12y – 28 = 0… + 19x + 38 = 0 19x = –38 x = –2 Sub x = –2 into 5(–2) + 6y – 14 = 0 –10 + 6y – 14 = 0 6y = 24 y = 4 Check in (–2, 4) 3(–2) – 4(4) + 22 = 0 – 6 – 16 + 22 = 0 True, (–2, 4)
Example 2 Show the lines 2x – 5y + 19 = 0, 5x + 3y – 30 = 0 and 6y – 5x – 15 = 0 are concurrent. Remember concurrent lines all pass through the same point. 2x – 5y + 19 = 0 … 5x + 3y – 30 = 0 … To show the lines are concurrent, sub the point into the third equation. × 3 6x – 15y + 57 = 0… × 5 25x + 15y – 150 = 0… + 31x – 93 = 0 31x = 93 x = 3 6(5) – 5(3) – 15 = 0 30 – 15 – 15 = 0 0 = 0 (3, 5) lies on this line The lines are concurrent. Sub x = 3 into 5(3) + 3y – 30 = 0 15 + 3y – 30 = 0 3y = 15 y = 5 Check in (3, 5) 2(3) – 5(5) + 19 = 0 6 – 25 + 19 = 0 True, (3, 5)
Today’s work Exercise 7.8 Page 294 Q2→5 Then “some” of Q1
Yesterday’s work Exercise 7.7 Page 291→292 Q1→4, 10 & 11 P 13cm