SOLVING LINEAR EQUATIONS WITH FRACTIONAL COEFFICIENTS

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To solve linear equations with rational coefficients, we have to know operations on fractions.

  • To add or subtract two or more fractions, we should make the denominators same. If the denominators are not same, we have to take least common multiple and simplify.
  • Inverse operation of plus is minus.
  • Inverse operation of minus is plus.
  • Inverse operation of multiplication is division.
  • Inverse operation of division is multiplication.

Solve each equation.

Problem 1 :

m + 4 = 13/2

Solution :

m + 4 = 13/2

Subtract 4 from each side.

m + 4 - 4 = (13/2) – 4

m = (13 – 8)/2

m = 5/2

Problem 2 :

8/3 = x – 1 1/3

Solution :

8/3 = x – 1 1/3

8/3 = x – 4/3

Adding 4/3 on both side.

8/3 + 4/3 = 4/3 + x – 4/3

12/3 = x

4 = x

Problem 3 :

4/5 + v = 41/20

Solution :

4/5 + v = 41/20

Subtract  4/5 from each side.

4/5 + v – 4/5 = 41/20 – 4/5

v = 41/20 – 4/5 × (4/4)

v = 41/20 – 16/20

v = 25/20

v = 5/4

Problem 4 :

-11/5 = -2 + n

Solution :

-11/5 = -2 + n

Adding 2 on each side.

-11/5 + 2 = -2 + n + 2

(-11 + 10)/5 = n

-1/5 = n

Problem 5 :

-17/4 = v - 2

Solution :

-17/4 = v - 2

Adding 2 on each side.

-17/4 + 2 = v – 2 + 2

(-17 + 8)/4 = v

-9/4 = v 

Problem 6 :

x + 1 = 11/5

Solution :

x + 1 = 11/5

Subtract 1 from each side.

x + 1 – 1 = 11/5 - 1

x = (11 – 5)/5

x = 6/5

Problem 7 :

(2/3)x = -1

Solution :

(2/3)x = -1

Multiplying both side by 3.

(2/3)x . 3 = -1.3

2x = -3

Divide each side by 2.

2x/2 = -3/2

x =-3/2

Problem 8 :

-3/2 = -3/2x

Solution :

-3/2 = (-3/2)x

Multiplying both side by 2.

(-3/2) 2 = (-3/2) . 2

-3 = -3x

Divide each side by 3.

-3/3 = -3x/3

-1 = -x

x = 1

Problem 9 :

(2/5)x = -1/10

Solution :

(2/5)x = -1/10

Using cross multiplication.

20x = -5

Diving by 20

x = -1/4

Problem 10 :

-9/4 = -5x/4

Solution :

-9/4 = -5x/4

Multiplying both side by 4.

(-9/4) . 4 = (-5x/4) . 4

-9 = -5x

Divide each side by -5.

-9/(-5) = -5x/(-5)

9/5 = x

Problem 11 :

After earning interest, the balance of an account is $420. The new balance is 7/6 of the original balance. How much interest was earned?

Solution :


Original balance = $420

New balance = (7/6) of original balance

= (7/6) of 420

= (7/6) 420

= 7 ⋅ 70

= 490

So, the interest earned is $490.

Problem 12 :

Find the value of the variable. Then find the angle measures of the polygon. Check the reasonableness of your answer.

solving-linear-equation-with-fra-coeff-q1

Solution :

Sum of interior angles of the polygon = 540

b + (3/2)b + (2b - 90) + 90 + (b + 45) = 540

b + 3b/2 + 2b - 90 + 90 + b + 45 = 540

b + 3b/2 + 2b + b = 540 - 45

4b + (3b/2) = 495

(8b + 3b)/2 = 495

11b / 2 = 495

11b = 495(2)

11b = 990

b = 990/11

b = 90

Problem 13 :

You order two servings of pancakes and a fruit cup. The cost of the fruit cup is $1.50. You leave a 15% tip. Your total bill is $11.50. How much does one serving of pancakes cost?

Solution :

Total bill = 11.50

Cost of fruit cup = $1.50

Cost of pancake + cost of fruit cup = 11.50

Let x be the cost of one pancake.

115% of (x + 1.50) = 11.50

1.15(2x + 1.50) = 11.50

2x + 1.50 = 11.50/1.15

2x + 1.50 = 10

2x = 10 - 1.50

2x = 8.5

x = 8.5/2

x = 4.25

Cost of one pancake is $4.25.

Problem 14 :

Find the value of x. Then find the angle measures of the polygon.

solving-linear-equation-with-fra-coeff-q2.png

Solution :

Sum of interior angles = 360

x + x - 35 + x - 46 + (1/2)x = 360

3x + (x/2) - 81 = 360

(6x + x)/2 = 360 + 81

7x/2 = 441

x = 441(2/7)

x = 126

So, the value of x is 126.

Problem 15 :

Use the table to write and solve an equation to find the number of points p you need to score in the fourth game so that the mean number of points is 20?

Game

1

2

3

4

Points

25

15

18

p

Solution :

Mean = 20

(25 + 15 + 18 + p)/4 = 20

58 + p = 20(4)

58 + p = 80

p = 80 - 58

p = 22

So, the value of p is 22.

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