SOLVING ONE STEP EQUATIONS

Solve each of the following equation.

Example 1 :

x + 14 = 21

Solution :

x + 14 = 21

Subtract 14 on both sides.

x = 21-14

x = 7

Example 2 :

-13 = -4+h

Solution :

-13 = -4+h

Add 4 on both sides, we get

-13 + 4 = h

h = -9

Example 3 :

5y = 30

Solution :

5y = 30

Divide by 5 on both sides.

y = 30/5

y = 6

Example 4 :

x/8 = -4

Solution :

x/8 = -4

Multiply by 8 on both sides.

x = -4(8)

x = -32

Example 5 :

John buys organic almonds priced at $77 from the grocery store. How much did he pay the cashier, if he received $23 in change?

Solution :

Cost of almond = $77

Let x be the amount he has.

x - 77 = 23

x = 23 + 77

x = $100

Example 6 :

Peter and Mia participated in a quiz contest. They scored 23 points in all. If Peter scored 9 points, how many points did Mia score?

Solution :

Let x be the score of Mia.

They together scored 23 points. Peter scored 9 points.

x + 9 = 23

Subtract 9 on both sides.

x = 23-9

x = 14

So, Mia scored 14 points.

Example 7 :

Grace ran 9.7 miles less than Perry last week. Grace ran 13.8 miles. How many miles did Perry run?

Solution :

Let x be the number of miles that Perry ran.

Grace ran x-9.7 miles.

Number of miles grace ran = 13.8

x-9.7 = 13.8

x = 13.8+9.7

x = 23.5

Example 8 :

Shreya is cooking bread. The recipe calls for 5 1/6 cups of flour. She has already put in 3 9/10 cups. How many more cups does she need to put in?

Solution :

Let x be quantity of flour she need to put.

x + 3  9/10 = 5  1/6

Converting the mixed fraction into improper fraction.

x + 39/10 = 31/6

Subtract 39/10 on both sides.

x = (31/6)-(39/10)

x = (155-117)/30

x  = 38/30

x = 19/15

x = 1  4/15

Example 9 :

After paying $4.17 for a sandwich, Kali has $15.70. With how much money did she start?

Solution :

Let x be the amount he has initially.

x-4.17 = 15.70

Add 4.17 on both sides, we get

x = 15.70+4.17

x = 19.87

So, he has $19.87 initially.

Example 10 :

The product of x and three is negative-27. What is x?

Solution :

Product of x and three is = -27

3x = -27

dividing by 3 on both sides, we get

x = -27/3

x = -9

Example 11 :

What value of k makes the equation k + 4 ÷ 0.2 = 5 true?

a) −15         b) −5        c)  −3        d) 1.5

Solution :

k + 4 ÷ 0.2 = 5

Dividing 4 b y 0.2, we get 20

k + 20 = 5

Subtracting 20 on both sides

k = 5 - 20

k = -15

So, option a is correct.

Example 12 :

The melting point of a solid is the temperature at which the solid becomes a liquid. The melting point of bromine is 1/30 of the melting point of nitrogen. Write and solve an equation to find the melting point of nitrogen

Solution :

Let n be the melting point of the nitrogen.

Melting point of bromine = 1/30 of n

From the given picture, the melting point of bromine = -7^o C

-7 = (1/30) of n

-7 = (1/30) n

-7 = n/30

Multiplying by 30 on both sides, we get

-7(30) = n

n = -210

So, the melting point of nitrogen is -210^o C.

Example 13 :

Solve p − 8 ÷ 1/2 = −3

Solution :

p − 8 ÷ 1/2 = −3

Dividing 8 by 1/2, we get

p − 8/(2/1) = −3

p - 16 = -3

Adding 16 on both sides

p = -3 + 16

p = 13

So, the value of p is 13.

Example 14 :

Solve q + ∣ −10 ∣= 2

Solution :

q + ∣ −10 ∣= 2

The value of ∣ −10 ∣ is 10.

q + 10 = 2

Subtracting 10 on both sides.

q = 2 - 10

q = -8

So, the value of q is -8.

Example 15 :

The melting point of mercury is about 1/4 of the melting point of krypton. The melting point of mercury is −39° C. Write and solve an equation to fi nd the melting point of krypton.

Solution :

Let k be the melting point of krypton.

Melting point of mercury = −39° C

Melting point of mercury = 1/4 of k

1/4 of k = -39

(1/4)k = -39

k/4 = -39

Multiplying by 4, we get

k = -39(4)

k = -156° C

So, the melting point of krypton is -156° C.