Use the substitution method to solve 1 4 x + 1 6 y = 1 and x − y = 3 . Check the solution graphically.

We have, 1 4 x + 1 6 y = 1 . . . 1 and x − y = 3 ⇒ x = 3 + y . . . 2 Putting the value of x from equation 2 in equation 1 , we get: 3 + y 4 + y 6 = 1 ⇒ 3 3 + y + 2 y 12 = 1 ⇒ 9 + 3 y + 2 y = 12 ⇒ 5 y = 12 - 9 ⇒ 5 y = 3 ⇒ y = 3 5 = 0 . 6 Putting the value of y in equation 1 , we get: x 4 + 1 6 × 3 5 = 1 ⇒ x 4 + 1 10 = 1 ⇒ x 4 = 1 - 1 10 ⇒ x 4 = 9 10 ⇒ x = 18 5 = 3 . 6 Verification: Making the table for x and y for the equation x 4 + y 6 = 1 x 0 4 y 6 0 Making the table for x and y for the equation x - y = 3 x 0 3 y - 3 0 Plotting the points on the graph, we get: From the graph we can observe that both the equations intersect at the point 3 . 6 , 0 . 6 . Hence, the required solution is 3 . 6 , 0 . 6 .

Use the elimination method to solve 1 3 x + 1 2 y = − 1 4 and 1 6 x − 5 6 y = 11 16 . Check the solution graphically.

We have, 1 3 x + 1 2 y = − 1 4 . . . 1 and x 6 − 5 y 6 = 11 16 . . . 2 Multiply equation 1 with 5 3 , we get: 5 x 9 + 5 y 6 = − 5 12 . . . 3 Adding equation 3 and 2 , we get: 5 x 9 + 5 y 6 + x 6 − 5 y 6 = − 5 12 + 11 16 ⇒ 5 x 9 + x 6 = − 5 12 + 11 16 ⇒ 10 x + 3 x 18 = - 20 + 33 48 ⇒ 13 x 18 = 13 48 ⇒ x = 18 48 ⇒ x = 3 8 = 0 . 375 Putting the value of x in equation 1 , we get: 1 3 × 3 8 + y 2 = − 1 4 ⇒ 1 8 + y 2 = − 1 4 ⇒ y 2 = − 1 4 - 1 8 ⇒ y 2 = - 2 - 1 8 = - 3 8 ⇒ y = - 3 4 = - 0 . 75 Verification: Making the table for x and y for the equation 1 3 x + 1 2 y = − 1 4 or 2 x + 3 y = - 3 2 x 0 - 3 4 y - 1 2 0 Making the table for x and y for the equation 1 6 x − 5 6 y = 11 16 or x - 5 y = 33 8 x 0 33 8 y - 33 40 0 Plotting the points on the graph, we get: From the graph we can observe that both the equations intersect at the point 0 . 375 , - 0 . 75 . Hence, the required solution is 0 . 375 , - 0 . 75 .

E M B I B E

MYP Mathematics A concept based approach 4&5 Standard>Equivalence>Chapter 22 - A Model of Equality (Equivalence Transformations)>Mixed practice

Rose Harrison and Clara Huizink Solutions for Exercise 22: Mixed practice

Author:Rose Harrison & Clara Huizink

Rose Harrison Mathematics Solutions for Exercise - Rose Harrison and Clara Huizink Solutions for Exercise 22: Mixed practice

Attempt the practice questions from Exercise 22: Mixed practice with hints and solutions to strengthen your understanding. MYP Mathematics A concept based approach 4&5 Standard solutions are prepared by Experienced Embibe Experts.

Questions from Rose Harrison and Clara Huizink Solutions for Exercise 22: Mixed practice with Hints & Solutions

EASY

MYP:4-5

IMPORTANT

MYP Mathematics A concept based approach 4&5 Standard>Chapter 22 - A Model of Equality (Equivalence Transformations)>Mixed practice>Q 1

Find the value of $x$ for the equation $4 - 3 x = 5 x - 3$ by identifying the equivalence transformations used at each step.

EASY

MYP:4-5

IMPORTANT

MYP Mathematics A concept based approach 4&5 Standard>Chapter 22 - A Model of Equality (Equivalence Transformations)>Mixed practice>Q 1

Solve the equation $3 (3 x - 1) = 4 (2 x - 3)$ by identifying the equivalence transformations used at each step.

MEDIUM

MYP:4-5

IMPORTANT

MYP Mathematics A concept based approach 4&5 Standard>Chapter 22 - A Model of Equality (Equivalence Transformations)>Mixed practice>Q 1

Solve the equation $\frac{2}{3} x + 2 = 4 - \frac{1}{2} x$ by identifying the equivalence transformations used at each step. (Write the answer in fraction.)

HARD

MYP:4-5

IMPORTANT

MYP Mathematics A concept based approach 4&5 Standard>Chapter 22 - A Model of Equality (Equivalence Transformations)>Mixed practice>Q 2

Use the substitution method to solve $y = 4 x + 3 and y = - x - 2$ . Check the solution graphically.

HARD

MYP:4-5

IMPORTANT

MYP Mathematics A concept based approach 4&5 Standard>Chapter 22 - A Model of Equality (Equivalence Transformations)>Mixed practice>Q 2

Use the substitution method to solve $x - 7 y = 19 and 5 x - 8 y = - 13$ . Check the solution graphically.

HARD

MYP:4-5

IMPORTANT

MYP Mathematics A concept based approach 4&5 Standard>Chapter 22 - A Model of Equality (Equivalence Transformations)>Mixed practice>Q 2

Use the substitution method to solve $\frac{1}{4} x + \frac{1}{6} y = 1 and x - y = 3$ . Check the solution graphically.

HARD

MYP:4-5

IMPORTANT

MYP Mathematics A concept based approach 4&5 Standard>Chapter 22 - A Model of Equality (Equivalence Transformations)>Mixed practice>Q 2

Use the elimination method to solve $\frac{1}{3} x + \frac{1}{2} y = - \frac{1}{4} and \frac{1}{6} x - \frac{5}{6} y = \frac{11}{16}$ . Check the solution graphically.