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Linear Equation can be defined as an equation that has two variables which when plotted in a graph gives a straight line. The linear equation in two variables contains two unknown variables. Variable is a term that is not a number and is used in place of a number. Thus the equation is a combination of numbers and variables. The linear equations in two variables are in this form ax+by+c = 0. Here a,b,c are real numbers, and x,y are the two unknown variables. We will be assuming, if x,y values are not given or inserting the given values of x and y values in order to plot them on the graph.
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Key Takeaways: Linear equation, Variable, real numbers, nonlinear equations, graph, number line.
Linear Equation in one variable
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Considering the equation 2x + 5 = 0. Solving for x, the answer will be -52. The given linear equation has one variable in it thus graphically, it can be plotted on the x-axis. Let’s use the number line for plotting,
Numberline Plotting
In this situation, the equation has one variable thus can be plotted in a number line but when an equation has two variables, it is to be plotted in an x-axis and y-axis using graphs.
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Linear Equation in Two Variables
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Considering the equation 2x + 3y – 12 = 0, it is in ax + by + c = 0 form. Comparing with the equation the values of a,b,c are 2,3,-12. Now the two unknown variables x,y can have any value as assumed. But here we can see that when we put (x,y) as (3,2) it satisfies 2x + 3y = 12. Similarly, (x,y) as (0,4), (6,0) also satisfies the equation. Apart from these solutions, we can also find different solutions by considering the value of any one variable. For example if we consider x to be 1 then solving the equation will give us-
2x + 3y = 12
= 2(1) + 3y = 12
= 2 + 3y = 12
= 3y = 12-2
= 3y = 10
→ y = 10/3
Thus here, (x,y) = (1, 10/3)
Thus linear equations in two variables can have many solutions.
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Graph of Linear Equation in Two Variables
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Graphical representation requires the plotting of the x and y values of the equation on the coordinate plane in graph paper. For plotting the graph you will at least require 3 sets of points. It is important that those points fall in a straight line. If the points are randomly placed, it indicates some fault in your work. Let’s consider-
x + 2y = 6
| Considering x as | 0 | 2 | 4 | 6 |
| We get y as | 3 | 2 | 1 | 0 |
| (x,y) | (0,3) | (2,2) | (4,1) | (6,0) |
Draw the x and y-axis on the graph paper. In simpler ways, the x and y-axis are two number lines placed perpendicular to each other at the (0,0) point. This point is known as the origin. Use the above values to plot the points. Join these points with a straight line.
Line AB is x + 2y = 6. After plotting the line we can observe that the line crosses many points apart from the ones that we plotted. All of those points are solutions to the equation. Any point apart from the line is not the solution of the equation.
The video below explains this:
Graph of a Linear Equation in Two Variables Detailed Video Explanation:
Also Read: Complex numbers and Quadratic equations
Graph for Line Passing through Origin
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In special cases when the equations have (0,0) as one of its solutions, the solution line tends to pass through the origin of the x and y-axis. The graph resulting from the equation will have a form like
» y = kx
» x + y = 0
» x = ky
They will surely have one solution as (0,0). Let’s see an example,
y = 3x
| Considering x as | 0 | 2 | -2 |
| We get y as | 0 | 6 | -6 |
| (x,y) | (0,0) | (2,6) | (-2, -6) |
Other examples of getting a graph with the solution line passing through the origin is 2x = 3y, 4x + y = 0, y = 2x, etc.
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Graph for Line Parallel to One Axis
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Equations with solution lines being parallel to one of the axes occur when one of the variables has ‘0’ as its solution. Here the equation has value for only one variable. The value of the known variable remains the same for all the sets of solutions while the value of the other unknown variable can be taken as any value. In the Cartesian plane, all the values on the x-axis have their y-coordinate values as zero (x, 0) and all the values on the y-axis have their x-coordinate values as zero (0, y).
For lines parallel to x- axis have this form of equation » 0 × x + y – a = 0 » y – a = 0 » y = a
For lines parallel to y- axis have this form of equation » x + 0 × y – a = 0 » x – a = 0 » x = a
Where ‘a’ can be any number. Let’s consider this equation 2x + 1 = x – 3 » x = – 4
Now all the values of y are permissible because 0 × y is always 0. However, x must satisfy the equation x = – 4. While representing it in a linear equation with one variable we can use a number line but in a linear equation with two-equation –
For equation y = 3 i.e. 0 × x + 1 × y = 3
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Things to Remember
- The first value in the set of solutions for an equation is always the value of x-coordinate thus the other value is of the y-coordinate.
- Linear equation in one variable has a unique solution whereas linear equation in two variables has infinitely many solutions
- The solution of a linear equation is not affected when the same number is added to or subtracted from both sides of the equation.
- The solution of a linear equation is not affected when the same non-zero number is multiplied or divided on both sides of the equation.
- Every point in the solution line which plots the equation in two variables on the graph shows the values satisfying the equation.
- The solution line for the equation in two variables is unique for every equation despite having infinite solution values. All of those values lie on that line.
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Sample Questions
Ques. Write each of the following in the form of ax + by + c = 0 [3 marks]
2x + 3y = 4.37
4 = 5x – 3
2x = y
Ans. a. 2x + 3y = 4.37 can be written as 2x + 3y – 4.37 = 0. Here a = 2, b = 3 and c = – 4.37.
- The equation 4 = 5x – 3y can be written as 5x – 3y – 4 = 0. Here a = 5, b = –3 and c = – 4. It can also be written as –5x + 3y + 4 = 0 In this case a = –5, b = 3 and c = 4.
- The equation 2x = y can be written as 2x – y + 0 = 0. Here a = 2, b = –1 and c = 0.
Ques. Write each of the following as an equation in two variables: [3 marks]
x = –5
y = 2
2x = 3
Ans. a) x = –5 can be written as 1x + 0y = –5, or 1x + 0y + 5 = 0.
- b) y = 2 can be written as 0x + 1y = 2, or 0x + 1y – 2 = 0.
- c) 2x = 3 can be written as 2x + 0y – 3 = 0.
Ques. The cost of a notebook is twice the cost of a pen. Write a linear equation in two variables to represent this statement. [2 marks]
Ans. Let the cost of a notebook be ‘x’ and that of a pen be ‘y’
As per the statement, ‘The cost of a notebook is twice the cost of a pen’,
Equation can be written as x = 2y.
Ques. Find two solutions for each of the following equations: [3 marks]
4x + 3y = 12
2x + 5y = 0
3y + 4 = 0
Ans. a) Taking x = 0, we get 3y = 12, i.e., y = 4. So, (0, 4) is a solution of the given equation. Similarly, by taking y = 0, we get x = 3. Thus, (3, 0) is also a solution.
- b) Taking x = 0, we get 5y = 0, i.e., y = 0. So (0, 0) is a solution of the given equation. Now, if you take y = 0, you again get (0, 0) as a solution, which is the same as the earlier one. To get another solution, take x = 1, say. Then you can check that the corresponding value of y is - 25. So (1, - 25) is another solution of 2x + 5y = 0.
- c) Writing the equation 3y + 4 = 0 as 0x + 3y + 4 = 0, you will find that y = - 43 for any value of x. Thus, two solutions can be given as (0, - 43) and (1, - 43)
Ques. Find the value of k, if x = 2, y = 1 is a solution of the equation 2x + 3y = k.[2 marks]
Ans. As given x = 2, y = 1, substitute it in the equation 2x + 3y = k
2 (2) + 3 (1) = k
4 + 3 = k
k = 7
Ques. Check which of the following are solutions of the equation x – 2y = 4 and which are not: [3 marks]
(0, 2)
(2, 0)
(4, 0)
Ans. Here we need to check if the LHS is equal to RHS or not
- (0,2) in x – 2y = 4
LHS
= 0 – 2(2)
= - 4
Which is not equal to RHS, thus it is not a solution of the equation
- (2,0) in x – 2y = 4
LHS
= 2 – 2(0)
= 2
Which is not equal to RHS, thus it is not a solution of the equation
- (4,0) in x – 2y = 4
LHS
= 4 – 2(0)
= 4
LHS = RHS
Thus it is the solution of the equation
Ques. Draw the graph of x + y = 7.[3 marks]
Ans.
| x | 0 | 7 |
| y | 7 | 0 |
| (x,y) | (0,7) | (7,0) |
Graph:
Ques. If the point (3, 4) lies on the graph of the equation 3y = ax + 7, find the value of a.[2 marks]
Ans. Substituting (3,4) in 3y = ax + 7
3(4) = a (3) + 7
12 = 3a + 7
12 – 7 = 3a
5 = 3a
a = 53
Ques. The taxi fare in a city is as follows: For the first kilometre, the fare is Rs 8 and for the subsequent distance it is Rs 5 per km. Taking the distance covered as x km and total fare as y, write a linear equation for this information, and draw its graph. [5 marks]
Ans. Total distance covered = x km
Fare for the first km = Rs 8
Fare for the rest of the distance = Rs 5 × remaining distance
= Rs 5 (x-1)
Thus total fare = fare for first km + fare for the rest (x-1) km
y = 8 + (x-1) 5
y = 8 + 5x – 5
y = 5x + 3
As we need two solutions to draw the graph,
Consider x as 0,
y = 5(0) + 3
y = 0 + 3
y + 3
(x,y) = (0,3)
Consider x as 1,
y = 5(1) + 3
y = 5 + 3
y = 8
(x,y) = (1,8)
| x | 0 | 1 |
| y | 3 | 8 |
| (x,y) | (0,3) | (1,8) |
Graph –
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