Tips And Tricks And Shortcuts on Linear Equation Problems

Tips And Tricks And Shortcuts On Linear Equations

Mastering linear equations is essential for excelling in mathematics, and having the right tips and tricks up your sleeve can make problem-solving a breeze. First, identify the form of the equation, whether it’s in standard form, slope-intercept form, or point-slope form. Simplify the equation by combining like terms and eliminating fractions if present. Use the graphical method to visualize the solution and find the point of intersection.

Tips And Tricks on Linear Equation

Linear Equation Tips and Tricks and Shortcuts

Tips and Tricks :

• Here, we have provided quick and easy tips and tricks for you on Linear Equation questions which and efficiently in competitive exams as well as other recruitment exams that must help to find a better place.

• It can be easily solved by eliminating the wrong options. It means put the given values in equation and check which one is satisfying the equation.

• Standard form of linear equations is y= mx+b. There are 2 types of questions asked in exams explain below.

Shortcuts for solving Linear Equation:

There are several shortcuts and techniques to solve linear equations quickly and efficiently. Here are some useful shortcuts:

  1. Combine Like Terms: Before solving, combine similar terms on both sides of the equation to simplify the equation.

  2. Isolate the Variable: Aim to get the variable (usually ) on one side of the equation by applying inverse operations (addition, subtraction, multiplication, division) to move all other terms to the other side.

  3. Use Fractional Form: When the equation contains fractions, it’s often easier to work with the equation in fractional form to avoid dealing with large numbers.

  4. Cancellation: If you have terms with the same factor on both sides of the equation, you can cancel them out to simplify the equation.

  5. Multiply to Eliminate Fractions: To eliminate fractions, multiply the entire equation by the least common multiple (LCM) of the denominators.

  6. Cross-Multiplication: For equations with proportions or fractions, use cross-multiplication to simplify and solve for the variable.

💡Using these shortcuts and techniques can save you time and improve your problem-solving skills for linear equations. However, it’s essential to remember the principles behind each shortcut to ensure accuracy in your solutions. 

Type 1: Linear Equations Shortcuts. To Find the value of x or y.

Question 1. If 3a + 6 = 4a − 2, then find the value of a?


A. 3

B. 8

C. 6

D. 7

Solution:    We can use the trick of eliminating the option

Option 1, put a = 3

3 * 3 + 6 = 15

4 * 3 -2 = 10

This means option 1 is incorrect.

Now, check for option 2, put a = 8

3 * 8 + 6 = 30

4 * 8 – 2 = 30

This means option 2 satisfies the equation. Therefore, it is the correct option.

Correct option: B

Question 2. Which of the following is the correct equation of the line passing through the points (2, 5) and (4, 11)?


A) y = 3x + 1

B) y = 2x + 1

C) y = 2x + 3

D) y = 3x + 5

Solution:  To find the equation of a line passing through two given points, we first calculate the slope (m) using the formula:

m = \frac{(y2-y1)}{(x2-x1)}

For the points (2, 5) and (4, 11), the slope is \frac{(11-5)}{(4-2)}= \frac{6}{2}=3.

Next, we use the point-slope form of the linear equation:

y – y1 = m(x – x1)

Plugging in the values (2, 5) and m = 3, we get the equation y – 5 = 3(x – 2). Solving for y, we find y = 3x + 1.

Correct Option : A

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Type 2: Tips And Tricks for Linear Questions Word problems

Question 3. The cost of 5 blankets and 6 bedsheets is Rs.1500. The cost of 6 blankets and 5 bedsheets is Rs.1300. Find out the total cost of one blanket and one bedsheet.


A. Rs. 255

B. Rs. 250

C. Rs. 81.81

D. Rs. 254.545

Solution:    Let the cost of blankets be x and the cost of bedsheets be y.

According to the question:

5x+ 6y= 1500…(1)

6x+ 5y=1300…(2)

Multiply Eq 1 by 5 and Eq 2 by 6,

we get.

25x+30y = 7500…(3)

36x+30y = 7800…(4)

Subtract equation (3) from equation (4)

11x = 300

x = \frac{300}{11}

5 ×  \frac{300}{11} +6y =1500

6y = 1500 – \frac{1500}{11}

6y = 1500(1- \frac{1}{11})

6y = 1500 ×  \frac{10}{11}

y = \frac{2500}{11}

Total cost = x+y

=> \frac{300}{11}+\frac{2500}{11}

= \frac{2800}{11} = 254.545

Correct option: D

Question 4: What is the x-intercept of the line represented by the equation 2x + 3y = 12?


A) 4

B) 6

C) 8

D) 12

Solution:    To find the x-intercept, we set y = 0 and solve for x.

Substituting y = 0 into the equation 2x + 3y = 12, we get 2x + 3(0) = 12. Simplifying, we find 2x = 12, and then x = 6.

Correct Option: B

Question 5:  If the line 3x – y = 5 is parallel to the line 2x + ky = 8, what is the value of k?


A) -3

B) -2

C) 2

D) 3

Solution:    Two lines are parallel if their slopes are equal.

The slope of the line 3x – y = 5 can be found by rearranging the equation in slope-intercept form (y = mx + b), where m is the slope.

So, 3x – y = 5 becomes y = 3x – 5, and the slope is 3.

The line 2x + ky = 8 can also be rearranged as y = -(2/k)x + 8/k, where the slope is -2/k. To make both slopes equal (3 and -2/k), we need -2/k = 3. Solving for k, we find k = -2/3.

Correct Option: B

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