Solve the simultaneous equations. x + 2 y = 8 | x + 2 | + y = 6

We need to solve the simultaneous equations x + 2 y = 8 . . . . i x + 2 + y = 6 . . . . i i ⇒ ± x + 2 + y = 6 ⇒ x + 2 + y = 6 or - x + 2 + y = 6 ⇒ x + y = 4 . . . . i i i or - x + y = 8 . . . . i v Now, we need to solve equations i , and i i i Subtracting equation i i i from equation i . we have y = 4 Substituting the value of y in equation i , we have x + 8 = 8 ⇒ x = 0 Thus, we have x = 0 , y = 4 Now, we need to solve equations i , and i v . Adding equations i , and i v we have 3 y = 16 ⇒ y = 16 3 Substituting the value of y in equation i , we have x + 2 × 16 3 = 8 ⇒ x = 8 - 32 3 = - 8 3 Thus, another pair of solution for the given equation is x = - 8 3 , y = 16 3 Hence, the solutions for given simultaneous equations are x = 0 , y = 4 and x = - 8 3 , y = 16 3 .

Solve the simultaneous equations. 3 x + y = 0 y = 2 x 2 - 5

We need to solve the simultaneous equations: 3 x + y = 0 . . . . i y = 2 x 2 - 5 ⇒ y = 2 x 2 - 5 . . . . i i or 2 x 2 - 5 = - y . . . . . i i i Now, from equations i , and i i we have - 3 x = 2 x 2 - 5 ⇒ 2 x 2 + 3 x - 5 = 0 ⇒ x = - 3 ± 9 - 4 2 - 5 4 = - 3 ± 49 4 = - 3 ± 7 4 ⇒ x = 1 or x = - 2 . 5 Checking for these solutions, Substituting x = 1 in equation i , we have 3 + y = 0 ⇒ y = - 3 , which is not possible as 2 x 2 - 5 cannot be negative. Now, substituting x = - 2 . 5 in equation i we have 3 - 2 . 5 + y = 0 ⇒ y = 7 . 5 Thus, we get x = - 2 . 5 and y = 7 . 5 . . . . i v Now, from equations i , and i i i , we have 2 x 2 - 5 = 3 x ⇒ 2 x 2 - 3 x - 5 = 0 ⇒ 2 x 2 - 5 x + 2 x - 5 = 0 ⇒ x 2 x - 5 + 1 2 x - 5 = 0 ⇒ 2 x - 5 x + 1 = 0 ⇒ x = 2 . 5 or x = - 1 Substituting the value x = 2 . 5 in equation i , we have y = - 3 × 2 . 5 = - 7 . 5 , which is not possible since 2 x 2 - 5 = y Now, substituting x = - 1 in equation i , we have y = - 3 - 1 = 3 Hence, another solution for the given equation is x = - 1 , y = 3 . . . . v Therefore, the solution of the given equation is x = - 1 , y = 3 and x = - 2 . 5 , y = 7 . 5

Solve the equation | 3 x - 2 y - 11 | + 2 31 - 8 x + 5 y = 0 .

To Solve: | 3 x - 2 y - 11 | + 2 31 - 8 x + 5 y = 0 . ⇒ 3 x - 2 y - 11 = - 2 31 - 8 x + 5 y . . . . . i We know that a x + b = c x + d ⇒ a x + b = c x + d or a x + b = - c x + d Thus, from equation i , we have 3 x - 2 y - 11 = 2 31 - 8 x + 5 y . . . . . i i or 3 x - 2 y - 11 = - 2 31 - 8 x + 5 y . . . . . i i i Adding equations i i , and i i i we have 6 x - 4 y - 22 = 0 ⇒ 6 x - 4 y = 22 ⇒ 3 x - 2 y = 11 . . . . . i v Subtracting equation i i i from equation i i , we have 4 31 - 8 x + 5 y = 0 ⇒ 31 - 8 x + 5 y = 0 ⇒ 8 x - 5 y = 31 . . . . . . v Thus, we need to solve equations i v , and v Multiplying both sides of equation i v by 5 , and multiplying both sides of equation v by 2 , we have 15 x - 10 y = 55 . . . . v i 16 x - 10 y = 62 . . . . v i i Subtracting equation v i i from equation v i we have - x = - 7 ⇒ x = 7 Substituting the value of x in equation v i , we have 15 7 - 10 y = 55 ⇒ 105 - 10 y = 55 ⇒ 10 y = 105 - 55 = 50 ⇒ y = 5 Hence, the required solutions of the equations are x = 7 and y = 5 .

E M B I B E

Cambridge International AS & A Level Mathematics : Pure Mathematics 2 & 3 Course Book>Chapter 1 - Algebra>EXERCISE 1A

Sue Pemberton, Julianne Hughes and, Julian Gilbey Solutions for Chapter: Algebra, Exercise 2: EXERCISE 1A

Author:Sue Pemberton, Julianne Hughes & Julian Gilbey

Sue Pemberton Mathematics Solutions for Exercise - Sue Pemberton, Julianne Hughes and, Julian Gilbey Solutions for Chapter: Algebra, Exercise 2: EXERCISE 1A

Attempt the free practice questions on Chapter 1: Algebra, Exercise 2: EXERCISE 1A with hints and solutions to strengthen your understanding. Cambridge International AS & A Level Mathematics : Pure Mathematics 2 & 3 Course Book solutions are prepared by Experienced Embibe Experts.