Solved algebra problems on linear equations in one variable are explained below with the detailed explanation.
Let’s once again recall the methods of solving linear equations in one variable.
● Read the linear problem carefully and note what is given in the question and what is required to find out.
● Denote the unknown by any variable as x, y, ……. (any variable)
● Translate the problem to the language of mathematics or mathematical statements.
● Form the linear equation in one variable using the conditions given in the problems.
● Solve the equation for the unknown.
● Verify to be sure whether the answer satisfies the conditions of the problem.
1. The sum of three consecutive multiples of 4 is 444. Find these multiples.
Solution:
If x is a multiple of 4, the next multiple is x + 4, next to this is x + 8.
Their sum = 444
According to the question,
x + (x + 4) + (x + 8) = 444
⇒ x + x + 4 + x + 8 = 444
⇒ x + x + x + 4 + 8 = 444
⇒ 3x + 12 = 444
⇒ 3x = 444  12
⇒ x = 432/3
⇒ x = 144
Therefore, x + 4 = 144 + 4 = 148
Therefore, x + 8  144 + 8 – 152
Therefore, the three consecutive multiples of 4 are 144, 148, 152.
2. The denominator of a rational number is greater than its numerator by 3. If the numerator is increased by 7 and the denominator is decreased by 1, the new number becomes 3/2. Find the original number.
Solution:
Let the numerator of a rational number = x
Then the denominator of a rational number = x + 3
When numerator is increased by 7, then new numerator = x + 7
When denominator is decreased by 1, then new denominator = x + 3  1
The new number formed = 3/2
According to the question,
(x + 7)/(x + 3  1) = 3/2
⇒ (x + 7)/(x + 2) = 3/2
⇒ 2(x + 7) = 3(x + 2)
⇒ 2x + 14 = 3x + 6
⇒ 3x  2x = 14  6
⇒ x = 8
The original number i.e., x/(x + 3) = 8/(8 + 3) = 8/11
3. The sum of the digits of a two digit number is 7. If the number formed by reversing the digits is less than the original number by 27, find the original number.
Solution:
Let the units digit of the original number be x.
Then the tens digit of the original number be 7  x
Then the number formed = 10(7  x) + x × 1
= 70  10x + x = 70  9x
On reversing the digits, the number formed
= 10 × x + (7  x) × 1
= 10x + 7  x = 9x + 7
According to the question,
New number = original number  27
⇒ 9x + 7 = 70  9x  27
⇒ 9x + 7 = 43  9x
⇒ 9x + 9x = 43 – 7
⇒ 18x = 36
⇒ x = 36/18
⇒ x = 2
Therefore, 7  x
= 7  2
= 5
The original number is 52
4. A motorboat goes downstream in river and covers a distance between two coastal towns in 5 hours. It covers this distance upstream in 6 hours. If the speed of the stream is 3 km/hr, find the speed of the boat in still water.
Solution:
Let the speed of the boat in still water = x km/hr.
Speed of the boat downstream = (x + 3) km/hr.
Time taken to cover the distance = 5 hrs
Therefore, distance covered in 5 hrs = (x + 3) × 5 (D = Speed × Time)
Speed of the boat upstream = (x  3) km/hr
Time taken to cover the distance = 6 hrs.
Therefore, distance covered in 6 hrs = 6(x  3)
Therefore, the distance between two coastal towns is fixed, i.e., same.
According to the question,
5(x + 3) = 6(x  3)
⇒ 5x + 15 = 6x  18
⇒ 5x  6x = 18 – 15
⇒ x = 33
⇒ x = 33
Required speed of the boat is 33 km/hr.
5. Divide 28 into two parts in such a way that 6/5 of one part is equal to 2/3 of the other.
Solution:
Let one part be x.
Then other part = 28  x
It is given 6/5 of one part = 2/3 of the other.
⇒ 6/5x = 2/3(28  x)
⇒ 3x/5 = 1/3(28  x)
⇒ 9x = 5(28  x)
⇒ 9x = 140  5x
⇒ 9x + 5x = 140
⇒ 14x = 140
⇒ x = 140/14
⇒ x = 10
Then the two parts are 10 and 28  10 = 18.
6. A total of $10000 is distributed among 150 persons as gift. A gift is either of $50 or $100. Find the number of gifts of each type.
Solution:
Total number of gifts = 150
Let the number of $50 is x
Then the number of gifts of $100 is (150  x)
Amount spent on x gifts of $50 = $ 50x
Amount spent on (150  x) gifts of $100 = $100(150  x)
Total amount spent for prizes = $10000
According to the question,
50x + 100 (150  x) = 10000
⇒ 50x + 15000  100x = 10000
⇒ 50x = 10000  15000
⇒ 50x = 5000
⇒ x = 5000/50
⇒ x = 100
⇒ 150  x = 150  100 = 50
Therefore, gifts of $50 are 100 and gifts of $100 are 50.
The above stepbystep examples demonstrate the solved problems on linear equations in one variable.
● Equations
How to Solve Linear Equations?
Problems on Linear Equations in One Variable
Word Problems on Linear Equations in One Variable
Practice Test on Linear Equations
Practice Test on Word Problems on Linear Equations
● Equations  Worksheets
Worksheet on Word Problems on Linear Equation
7th Grade Math Problems
8th Grade Math Practice
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