Sec 3 G1 Maths Graphs of linear equations worksheet with answers

A free Sec 3 G1 Maths worksheet on graphs of linear equations: 7 questions, 30 marks, short questions first and then structured ones, in the school exam style. The answers and full worked solutions are on the last page.

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Sec 3 G1 Maths worksheet: Graphs of linear equations

Total: 30 marksSuggested time: 50 minutes

Name: ______________________________Class: ____________ Date: ____________

Answer all questions. Show all your working. Give non-exact answers correct to 3 significant figures, or 1 decimal place for angles in degrees, unless the question says otherwise.

Section A

State whether the gradient of the line shown is positive, negative, zero or not defined.

Answer[1]

The line passes through the points

A

and

B

shown. Find its gradient.

Answer[2]

Complete the table of values for

y = 2x + 4

Answer[3]

(a)Complete the table of values for

y = x + 1

Answer (a)[3]

(b)Complete the table of values for

4x + y = -5

Answer (b)[3]

(a)Complete the table of values for

y = 4x - 4

Answer (a)[3]

(b)Complete the table of values for

y = -x + 1

Answer (b)[3]

Section B

(a)State whether the gradient of the line shown is positive, negative, zero or not defined.

Answer (a)[1]

(b)(i)Find the

y

-intercept of the line

y = 6 - 4x

Answer (b)(i)[1]

(b)(ii)Find the gradient of the line

y = \dfrac{4}{3}x - 5

Answer (b)(ii)[1]

(c)Complete the table of values for

y = -2x + 5

Answer (c)[3]

(a)Complete the table of values for

3x + y = -1

Answer (a)[3]

(b)(i)Find the

y

-intercept of the line

2x - 4y = 11

Answer (b)(i)[2]

(b)(ii)State whether the gradient of the line

y = -1

is positive, negative, zero or not defined.

Answer (b)(ii)[1]

END OF PAPER

Answers and worked solutions

Question 1

Answer: NEGATIVE[1]

The line goes down from left to right.So the gradient is negative.

Question 2

Answer:

m = \dfrac{5}{3}

[2]

A(0, 0)

and

B(3, 5)

Rise:

5 - 0 = 5

Run:

3 - 0 = 3

m = \dfrac{5}{3}

Question 3

Answer:

y = 8

[3]

When

x = -3

y = 2(-3) + 4

y = -2

When

x = 1

y = 2(1) + 4

y = 6

When

x = 2

y = 2(2) + 4

y = 8

Question 4

(a)Answer:

y = 4

[3]

When

x = -2

y = (-2) + 1

y = -1

When

x = -1

y = (-1) + 1

y = 0

When

x = 3

y = (3) + 1

y = 4

(b)Answer:

y = -13

[3]

y = -4x - 5

When

x = -2

y = -4(-2) - 5

y = 3

When

x = 1

y = -4(1) - 5

y = -9

When

x = 2

y = -4(2) - 5

y = -13

Question 5

(a)Answer:

y = 0

[3]

When

x = -1

y = 4(-1) - 4

y = -8

When

x = 0

y = 4(0) - 4

y = -4

When

x = 1

y = 4(1) - 4

y = 0

(b)Answer:

y = -1

[3]

When

x = -1

y = -(-1) + 1

y = 2

When

x = 0

y = -(0) + 1

y = 1

When

x = 2

y = -(2) + 1

y = -1

Question 6

(a)Answer: NEGATIVE[1]

The line goes down from left to right.So the gradient is negative.

(b)(i)Answer:

c = 6

[1]

y = -4x + 6

c = 6

(b)(ii)Answer:

m = \dfrac{4}{3}

[1]

m = \dfrac{4}{3}

(c)Answer:

y = 7

[3]

When

x = -3

y = -2(-3) + 5

y = 11

When

x = -2

y = -2(-2) + 5

y = 9

When

x = -1

y = -2(-1) + 5

y = 7

Question 7

(a)Answer:

y = -1

[3]

y = -3x - 1

When

x = -2

y = -3(-2) - 1

y = 5

When

x = -1

y = -3(-1) - 1

y = 2

When

x = 0

y = -3(0) - 1

y = -1

(b)(i)Answer:

c = -\dfrac{11}{4}

[2]

-4y = -2x + 11

y = \dfrac{1}{2}x - \dfrac{11}{4}

c = -\dfrac{11}{4}

(b)(ii)Answer: ZERO[1]

The line goes flat (horizontal).So the gradient is zero.

Sec 3 G1 Maths Graphs of linear equations worksheet with answers

In the print window, choose Save as PDF to download it.

Or practise it online with step-by-step help when you get one wrong.

Section B

(a)State whether the gradient of the line shown is positive, negative, zero or not defined.

Answer (a)[1]

(b)(i)Find the

y

-intercept of the line

y = 6 - 4x

Answer (b)(i)[1]

(b)(ii)Find the gradient of the line

y = \dfrac{4}{3}x - 5

Answer (b)(ii)[1]

(c)Complete the table of values for

y = -2x + 5

Answer (c)[3]

Answers and worked solutions

Question 1

Answer: NEGATIVE[1]

The line goes down from left to right.So the gradient is negative.

Question 2

Answer:

m = \dfrac{5}{3}

[2]

A(0, 0)

and

B(3, 5)

Rise:

5 - 0 = 5

Run:

3 - 0 = 3

m = \dfrac{5}{3}

Question 3

Answer:

y = 8

[3]

When

x = -3

y = 2(-3) + 4

y = -2

When

x = 1

y = 2(1) + 4

y = 6

When

x = 2

y = 2(2) + 4

y = 8

Question 4

(a)Answer:

y = 4

[3]

When

x = -2

y = (-2) + 1

y = -1

When

x = -1

y = (-1) + 1

y = 0

When

x = 3

y = (3) + 1

y = 4

(b)Answer:

y = -13

[3]

y = -4x - 5

When

x = -2

y = -4(-2) - 5

y = 3

When

x = 1

y = -4(1) - 5

y = -9

When

x = 2

y = -4(2) - 5

y = -13

Question 5

(a)Answer:

y = 0

[3]

When

x = -1

y = 4(-1) - 4

y = -8

When

x = 0

y = 4(0) - 4

y = -4

When

x = 1

y = 4(1) - 4

y = 0

(b)Answer:

y = -1

[3]

When

x = -1

y = -(-1) + 1

y = 2

When

x = 0

y = -(0) + 1

y = 1

When

x = 2

y = -(2) + 1

y = -1

Question 6

(a)Answer: NEGATIVE[1]

The line goes down from left to right.So the gradient is negative.

(b)(i)Answer:

c = 6

[1]

y = -4x + 6

c = 6

(b)(ii)Answer:

m = \dfrac{4}{3}

[1]

m = \dfrac{4}{3}

(c)Answer:

y = 7

[3]

When

x = -3

y = -2(-3) + 5

y = 11

When

x = -2

y = -2(-2) + 5

y = 9

When

x = -1

y = -2(-1) + 5

y = 7

Question 7

(a)Answer:

y = -1

[3]

y = -3x - 1

When

x = -2

y = -3(-2) - 1

y = 5

When

x = -1

y = -3(-1) - 1

y = 2

When

x = 0

y = -3(0) - 1

y = -1

(b)(i)Answer:

c = -\dfrac{11}{4}

[2]

-4y = -2x + 11

y = \dfrac{1}{2}x - \dfrac{11}{4}

c = -\dfrac{11}{4}

(b)(ii)Answer: ZERO[1]

The line goes flat (horizontal).So the gradient is zero.