An integral part of the class 10 maths syllabus, Linear equations in two variables demonstrates the use of special algebraic equations. This chapter lays the foundation for the upcoming complex calculations. Thus, with a sound understanding of solving a linear equation with two variables, you will be able to quickly solve the advanced level problems. Just as we have helped you with class 9 linear equations in two variables, here we are with comprehensive notes for linear equations in two variables class 10. So, let’s begin with the blog and ace this topic.
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When is an Equation Considered to be a Linear Equation of Two Variables?
According to the Linear Equations in Two Variables CLass 10 chapter, a simple equation is considered to be a linear equation of two variables when it is presented in the form ax+by+c=0. In this case, a,b, and c should be constants, or real numbers, such as 2, 3, 4, etc. The coefficients of x and y that are in this case, a and b, should not be equal to 0. Here, the two unknown variables would be x and y, as a, b and c are real numbers. For example, 4x+5y+3=12 would be considered a linear equation of two variables.
The two unknown variables are x and y. Their respective coefficients, 4 and 5, are real numbers not equal to 0. Thus, to solve this equation, we will get two values, one for x and one for y, which will make the two sides of the equation equal.
Pair of Linear Equations in Two Variables PDF
Solution of the Linear Equation Of Two Variables
As per the Linear Equations in Two Variables class 10 chapter, the solution of a linear equation of two variables, such as ax+by=c, will be represented by a particular point on a graph. In this case, the x coordinate will be multiplied by a, to give the first value, and the coordinate will be multiplied by b, to provide a second value. The sum of these two values will be c.Thus, there is an infinite number of values and solutions in linear equations with two variables.
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Finding the Values of Variables
According to the Linear Equations in Two Variables class 10 chapter, the easiest and most common way of finding the values of the variables in linear equations with two variables is using the substitution method. In this method, we will substitute each of the variables, both x and y with 0.
For this equation, if we start by substituting x with 0, we get,
4(0)+ 5y= 30
Therefore, solving the equation,
we get y=6
We now substitute y with 0, to find the value of x
4x+ 5(0)= 30
Therefore, solving the equation, we get the value of x as 7.5.
This is the simplest method mentioned in the linear equation with two variables class 10 chapter.
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How to Solve Linear Equations with Two Variables Graphically?[optin-monster-shortcode id=”xf2mlnjiouddzrshykdb”]
There are several ways to solve linear equations with two variables. Yet, using a graph to plot them is one of the easiest and most accurate methods. It also provides an excellent visual representation of the equations to help students understand the concept better, especially for complex problems. To be able to solve linear equations with two variables graphically, we need to be able to identify two separate equations, each having two variables.
4x +5y=30- Equation 1
6x – 3y= 12- Equation 2
We now need to bring any one of these variables, x or y, to a common denominator, or the same number, such as 12x, (common for both 4 and 6) or 15y, (common for both 5 and 3). Then, we can subtract one equation from the other, and cancel this particular variable. The remaining equation will then have only one remaining variable, which will make it easy for us to find its value. Then, in the original equation, we can substitute this value and find the correct value of the other variable (the one we eliminated). Hence, using this method from the linear equations in two variables class 10 chapter, we can plot both the equations on a graph.
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Below are the steps to solve the above two equations:
Step 1: Multiplying both sides of Equation 1 by 3
3 (4x+5y)= 3(30)
This gives us:
12x + 15y= 90- Equation 3
Step 2: Multiplying both sides of Equation 2 by 2
This gives us:
12x-6y=24- Equation 4
Step 3: Subtracting Equation 4 from Equation 3
We get, 21y=66, which gives us the value of y as 3.14.
Step 4: We can substitute this value of y as 3.14 in Equation 1
This gives us:
4x + 5(3.14)= 30
Step 5: Thus, solving the equation, we get x = 3.57
Therefore y=3.14, and x= 3.57, in this case.
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Practice Questions for Linear Equations with Two Variables
- If x = 3m –1 and y = 4 is a solution of the equation x + y = 6, then find the value of m. [Answer: m=1]
- Akhila goes to a fair with 20 and wants to have rides on the Giant Wheel and play Hoopla. Represent this situation algebraically and graphically (geometrically).
- What is the point of intersection of the line represented by 3x – 2y = 6 and the y-axis. [Answer: (0, -3)]
- Romila went to a stationery shop and purchased 2 pencils and 3 erasers for 9. Her friend Sonali saw the new variety of pencils and erasers with Romila, and she also bought 4 pencils and 6 erasers of the same kind for 18. Represent this situation algebraically and graphically.
- In a deer park, the number of heads and number of legs of deer and human visitors were counted and it was found that there were 39 heads and 132 legs. Find the number of deer and human visitors in the park. [Answer: Dear: 27, Visitors: 12]
- For what value of p, system of equations 2x + py = 8 and x + y = 6 have no solution. [Answer: p=2]
- Form the pair of linear equations in the following problems, and find their solutions graphically.
- 10 students of Class X took part in a Mathematics quiz. If the number of girls is 4 more than the number of boys, find the number of boys and girls who took part in the quiz.
- 5 pencils and 7 pens together cost ` 50, whereas 7 pencils and 5 pens together cost ` 46. Find the cost of one pencil and that of one pen.
- Aftab tells his daughter, “Seven years ago, I was seven times as old as you were then. Also, three years from now, I shall be three times as old as you will be.” (Isn’t this interesting?) Represent this situation algebraically and graphically.
- The coach of a cricket team buys 3 bats and 6 balls for ` 3900. Later, she buys another bat and 3 more balls of the same kind for 1300. Represent this situation algebraically and geometrically.
- The cost of 2 kg of apples and 1kg of grapes on a day was found to be ` 160. After a month, the cost of 4 kg of apples and 2 kg of grapes is ` 300. Represent the situation algebraically and geometrically
- From Delhi station if we buy 2 tickets to station A and 3 tickets to station B, the total cost is ₹77, but if we buy 3 tickets to station A and 5 tickets to station B, the total cost is ₹124. What are fares from Delhi to station A and to station B? [Answer: ₹13, ₹17]
- A motor cyclist is moving along the line x – y = 2 and another motor cyclist is moving along the line x – y = 4 find out their moving direction. [Answer: move parallel]
- A farmer sold a calf and a cow for ₹760, thereby, making a profit of 25% on the calf and 10% on the cow. By selling them for ₹767.50, he would have realised a profit of 10% on the calf and 25% on the cow. Find the cost of each. [Answer: Cost of cow = ₹350, cost of calf = ₹300]
- Draw the graphs of the equations x – y + 1 = 0 and 3x + 2y – 12 = 0. Determine the coordinates of the vertices of the triangle formed by these lines and the x-axis, and shade the triangular region.
- Half the perimeter of a rectangular garden, whose length is 4 m more than its width, is 36 m. Find the dimensions of the garden.
- Solve the following pair of equations by substitution method: 7x – 15y = 2 (1) x + 2y = 3.
- Let us consider Example 2 in Section 3.3, i.e., the cost of 2 pencils and
- 3 erasers is ` 9 and the cost of 4 pencils and 6 erasers is ` 18. Find the cost of each pencil and each eraser
Hopefully, through our comprehensive blog about linear equations in two variables class 10, you now are familiar with the concept. For assistance regarding crucial decisions like choosing the right stream after class 10th, reach out to our experts at Leverage Edu.