Introduction to word problem involving linear equation:
In mathematics, word problems are given importance because it trains our mind to apply mathematics in real life situations.
Also, in real life, we only come across simple relations between known and unknown quantities. This type of relation, in mathematics, is called a linear equation. A linear equation establishes a relation between one or more variables and numbers.
Example of a Word Problem Based on Linear Equation:
In a word problem, it is very important to correctly transform verbal statements into algebraic equation and solve them mathematically to find the required information. Let us consider the following example.
Bob, John and Ann are three children. The sum the ages of all the children is 12. The difference of age between John and the sum of the ages of other two is 4. John is elder to Ann by 2 years.
With the above information find the ages of all the children.
Let B, J and A be the ages of Bob, John and Ann. Let us transform each statement into linear equations.
The sum the ages of all the children is 12.
This means, B + J + A = 12 ----(1)
The difference of age between John and the sum of the ages of other two is 4
This means, B – (J + A) = 4 -----(2)
John is elder to Ann by 2 years.
This means, J = A + 2 ------(3)
Now adding (1) and (2) gives, 2B = 16 or B = 8.
Plugging in the value of B in (1) or (2), gives J + A = 4 and therefore , J = -A + 4 -----(4)
Adding (3) and (4), you get 2J = 6 or J = 3
Plugging the values of B and J in (1), you can find A = 1
Hence the age of Bob is 8, the age of John is 3 and the age of Ann is 1.Looking out for more help on algebra word problems in algebra by visiting listed websites.
A Word Problem Involving Linear Equation in Real Life:
How many liters of 80% strong acid should be added to 5 liter of 20% strong acid to make a mixture of 60% strong acid?
It may seem to be a difficult problem. But actually it is simple if a linear equation is made by assuming V to be the volume to be added. The equation is,
5(0.2) + V(0.8) = (V + 5)(0.6) .
You can easily solve the above equation and figure out as V = 10 liters!
In mathematics, word problems are given importance because it trains our mind to apply mathematics in real life situations.
Also, in real life, we only come across simple relations between known and unknown quantities. This type of relation, in mathematics, is called a linear equation. A linear equation establishes a relation between one or more variables and numbers.
Example of a Word Problem Based on Linear Equation:
In a word problem, it is very important to correctly transform verbal statements into algebraic equation and solve them mathematically to find the required information. Let us consider the following example.
Bob, John and Ann are three children. The sum the ages of all the children is 12. The difference of age between John and the sum of the ages of other two is 4. John is elder to Ann by 2 years.
With the above information find the ages of all the children.
Let B, J and A be the ages of Bob, John and Ann. Let us transform each statement into linear equations.
The sum the ages of all the children is 12.
This means, B + J + A = 12 ----(1)
The difference of age between John and the sum of the ages of other two is 4
This means, B – (J + A) = 4 -----(2)
John is elder to Ann by 2 years.
This means, J = A + 2 ------(3)
Now adding (1) and (2) gives, 2B = 16 or B = 8.
Plugging in the value of B in (1) or (2), gives J + A = 4 and therefore , J = -A + 4 -----(4)
Adding (3) and (4), you get 2J = 6 or J = 3
Plugging the values of B and J in (1), you can find A = 1
Hence the age of Bob is 8, the age of John is 3 and the age of Ann is 1.Looking out for more help on algebra word problems in algebra by visiting listed websites.
A Word Problem Involving Linear Equation in Real Life:
How many liters of 80% strong acid should be added to 5 liter of 20% strong acid to make a mixture of 60% strong acid?
It may seem to be a difficult problem. But actually it is simple if a linear equation is made by assuming V to be the volume to be added. The equation is,
5(0.2) + V(0.8) = (V + 5)(0.6) .
You can easily solve the above equation and figure out as V = 10 liters!
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