Linear Equation

Linear Equation

8th Grade

5 Qs

quiz-placeholder

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Linear Equation

Linear Equation

Assessment

Quiz

Mathematics

8th Grade

Easy

Created by

MUHAMMAD FAIZ BIN SHAIRUL AZMAN STUDENT

Used 1+ times

FREE Resource

5 questions

Show all answers

1.

OPEN ENDED QUESTION

3 mins • 1 pt

Verify that x = 4 is the root of the equation 3x/2 = 6

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Answer explanation

To verify whether the given root is the solution of the given equation, substitute x = 4 in the equation 3x/2 = 6.

⇒ (3(4))/2 = 6

⇒ (12/2) = 6

⇒ 6 = 6

Hence, x = 4 is the root of the equation 3x/2 = 6.

2.

OPEN ENDED QUESTION

3 mins • 1 pt

If x = 2, then 2x – 5 = 7. Check whether the statement is true or false?

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Answer explanation

Given equation: 2x – 5 = 7

If x = 2,

= 2(2) – 5

= 4 – 5 = -1

Hence, the given statement is false

3.

OPEN ENDED QUESTION

3 mins • 1 pt

Express the equation x = 3y in the form of ax+by+c = 0 and find the values of a, b and c ?

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Answer explanation

Given equation: x = 3y

We know that the standard form of linear equation in two variables is ax+by+c = 0 …(1)

Now, rearranging the given equation, we get

⇒ x – 3y = 0

This can be written as

⇒ 1(x) + (-3)y + (0)c = 0 …(2)

On comparing equation (1) and (2), we get

⇒ a = 1, b = -3 and c = 0.

4.

OPEN ENDED QUESTION

3 mins • 1 pt

Find three solutions for the equation 2x + y = 7.

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Answer explanation

To find the solutions for the equation 2x + y = 7, substitute different values for x.

When x = 0,

⇒ 2(0) + y = 7

⇒ y = 7

Therefore, the solution is (0, 7).

When x = 1,

⇒ 2(1) + y = 7

⇒ y = 7 – 2

⇒ y = 5

Hence, the solution is (1, 5).

When x = 2,

⇒ 2(2) + y = 7

⇒ 4 + y = 7

⇒ y = 3

Hence, the solution is (2, 3).

Therefore, the three solutions are (0, 7), (1, 5) and (2, 3).

5.

OPEN ENDED QUESTION

3 mins • 1 pt

Solve the following equations using the substitution method:

3x + 4y = 10 and 2x – 2y = 2

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Answer explanation

3x + 4y = 10 …(1)

2x – 2y = 2 …(2)

Equation (2) can be written as:

2(x – y) = 2

x – y = 1

x = 1+y …(3)

Now, substitute (3) in (1), we get

3 (1+y) + 4y = 10

3 + 3y + 4y = 10

7y = 10 – 3

7y = 7

Hence, y = 1.

Now, substitute y = 1 in (3), we get

x = 1 + 1

x = 2.

Hence, x = 2 and y = 1 are the solutions of the given equations.