Math is useful across all fields and linear equations have proven useful in translating real word problems into numbers and solving for the unknown. In the presentation, I will try to explain what a linear equation is and how various components in a linear equation behaves. Finally, we will see examples of real problems and how to convert them into a linear equation.
A linear equation is a mathematical statement that usually contains not more than two variables denoted with any letters of the English alphabet such as x and y, and also a constant too(for example k). Another thing about linear equations is that the power of the variables is always less than or equal to one. Any equations that are raised to a power of 2 or more is no longer a linear equation. The unknown variables of a linear equation can be plotted in a graph - which is usually a straight line graph.
A linear equations is usually of the form below:
cx + d = y
The variables in this linear equation are x and y, while c and d remain the constants.
Solving Linear Equations: Solving linear equations is not as hard as many students feel. You just need a basic knowledge of arithmetic and you are good to go.
Here are some of the frequent steps you can take to solve linear equations. Just three of them below:
Linear equations can be useful when trying to determine the relationship between two variables. In order words, a real word problem could be expressed as a linear equation and then the solution could be found very quickly.
To convert real word problems to linear equations is simple, you just need to think critically. First try to evaluate the problem and express its parts using alphabets. Then perform basic math on them. We can understand these steps better with examples.
Lets see some examples.
Example 1: John is three times older than James. In ten years time, he will be twice as old as James. Find John's current age.
Solution.
Stage 1: We will convert the word problem into a linear equations using letters to represent constants and variables.
**If James age is a, then John's age is 3 times a = 3a. In 10 years time, the ages will be:
John = 3a + 10
James = a + 10
But John is twice as old as James in that 10 years time, so adding that, the final linear equation will be:
3a + 10 = 2(a + 10)
The above is the linear equation for that problem and it is easy to solve now.
Stage 2: Solving the linear equation
Here are the steps:
3a + 10 = 2(a + 10)
3a + 10 = 2a + 20 (We used 2 to multiply (a + 10) to remove the bracket)
3a - 2a = 20 - 10 (We collected like terms. Their signs change as they pass over the = symbol. + becomes - and vice versa).
a = 10 (We solved each side of the equation)
So James age is 10 years. But we are looking for John's age. Since James is 3 times older than John, then
John's age = 3 x 10 = 30 years.
Example 2: Mary and Jane are two friends. They shared $270 such that Mary got $50 more than Jane. How much did each friend get?
Solution
Stage 1 We need to convert the problem into a linear equation.
Let us represent Janes share with b.
Then Mary's share is b + 50
Their shares added together gives the total amount shared. So the final equation is:
b + b + 50 = 270
Now that we have arrived at the Linear equation, we can quickly find the unknowns.
b + b + 50 = 270
2b + 50 = 270 (We add the b's which are common)
2b = 270 - 50 (We collected like terms. As 50 crosses the = sign, it turned negative)
2b = 220 (We did a basic subtraction of the right side)
b=220/2 (Collected like terms again)
b = 110
Janes share (represented by b) is $110. But Mary's share is bigger than Jane's share by $50.
So Mary's share is: 110 + 50 = $160.
So out of the $270, Mary got $160 and Jane got $110.
Linear equations are cool for finding the relationship between two variables. The above two examples are real word problems which are converted to linear equations and their solution found. You can always use the same method to solve for problems like the one above.
