Question

1)Write the following statement in the form of an equation: If you subtract 3 from 6 times a number, you

(a) 3x – 6 = 9
(b) 6x – 3 = 9
(c) 6x + 3 = 9
(d) 3x + 6 = 9


2)Write the following statement in the form of an equation: Taking away 5 from x gives 10

(a) x – 5 = 10
(b) x + 5 = 10
(c) x – 10 = 5
(d) none of these


3)The solution of the equation x + 3 = 0 is

(a) -3
(b) 0
(c) 1
(d) 3


4)The solution of the equation x – 6 = 1 is

(a) 1
(b) 6
(c) -7
(b) 7
​

Answer 1

hope this helps u

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1) If you subtract 3 from 6 times a number

6x-3=9

2)Taking away 5 from x gives 10

x-5=10

3) x + 3 = 0

x = 0-3

x = -3

$\boxed{\red{x=\:-3}}$

4) x-6=1

x = 1+6

$\boxed{\red{x=7}}$

Answer 2

Given :   Taking away 5 from x gives 10

To Find : Write the  statement in the form of an equation

Solution:

Taking away 5 from x gives 10

Taking away means subtraction

Taking away 5 from x  means  x - 5

gives 10

=> x - 5 = 10

subtract 3 from 6 times a number gives 9

Let say number is x

6 times a number = 6x

subtract 3 from 6 times a number

6x - 3

subtract 3 from 6 times a number gives 9

6x - 3 = 9

x + 3 = 0

Subtract 3 from both sides

=> x + 3 - 3  = 0 - 3

=> x = - 3

x – 6 = 1

Add 6 both sides

=> x - 6 + 6 = 1 + 6

=> x  = 7

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