Content Curator
In this chapter, we will get to learn what is pair of Linear Equations in Two Variables, its formulas, and some questions to practice. A mathematical method of expressing a correlation between two variables is known as an equation. When an equation is represented in the form ax + by + c = 0, where a, b, c are real numbers, and a ≠ 0 and b ≠ 0, is called a linear equation with two variables. There are a few important key points to remember about these equations, viz.
- These equations have infinite solutions and every solution is a point of the graph.
- This equation represents a straight line on the graph
- If x = 0, then the equation represents y-axis, similarly if y = 0 then then equation represents x-axis
- On the other hand, in the graph of the line parallel to the y-axis, x = a, and for the line parallel to x-axis, y = b
- The equation of the line that passes through the origin is represented in the form y=mx
Pair of Linear equations in two variables will either have infinite solutions, a unique solution, or no solution depending on whether the lines are parallel, intersecting, or are coincident. Now that we know what these equations are let’s see how can we derive the solution to these equations algebraically.
First, let’s try to see the graphical method to find the solution to these linear equations. There are three possibilities to that, first, if the equation has no solution, then that pair is called an inconsistent pair of linear equations. The second possibility is that the equation will have a solution, such pairs of linear equations are termed as consistent pairs. Thirdly if the equation has infinite solutions, then that pair of equations will be termed as dependent pair of linear equations. Depending on the correlation between the coefficients the graphical representation of the equation will vary.
-
Intersecting lines:
The equation to intersecting lines will be in the form,
a1x+ b1y + c1=0
a2x+ b2y+c2=0
where,
a1a2 ≠ b1/b2
In this condition, the linear equation will have only one unique solution which will be the intersecting point coordinate, and will be represented graphically like this,
-
Coincident lines:
The equation to coincident lines will be in the form,
a1x+ b1y + c1=0
a2x+ b2y+c2=0
where,
a1a2 = b1b2 = c1c2
In this condition, the linear equation will have infinite solutions which will be any coordinate on the solution line and will be represented graphically like this,
-
Parallel Lines:
The equation to parallel lines will be in the form,
a1x+ b1y + c1=0
a2x+ b2y+c2=0
where,
a1a2 = b1b2 ≠ c1c2
In this condition, the linear equation will have no solution as the lines are parallel and will be represented graphically like this,
To find the solution to these pairs of equations algebraically there are three methods namely, Substitution method, Elimination method, and Cross Multiplication method. Let’s have a look at these methods in detail.
-
Substitution method:
Let’s consider the equations to be a1x+ b1y + c1=0 and a2x+ b2y+c2=0
Working of the method will be in the following steps:
- From one of the equations find the value of one of the variables say x in other variable term.
- Now that we have the value of one variable, substitute that value in the second equation to get an equation in x.
- Furthermore, solve the equation to get the value of x
- Now lastly, substitute this value of x in the equation for y that we derived in the first step and determine the value of y.
-
Elimination method:
Let’s consider the equations to be a1x+ b1y + c1=0 and a2x+ b2y+c2=0
Working of the method will be in the following steps:
- To begin with, make the coefficients of one of the variables numerically equal this can be derived by multiplying these coefficients by suitable constants.
- Now that one of the variables is equal, we can add or subtract the derived equations as the terms having similar coefficients will are having opposite or same signs, and thus, we can get an equation in only one variable as the result.
- After finding the equation, solve the same to get the value of one of the variables.
- Now lastly, substitute this value that we found in the above step in either of the two given equations in the problem and solve it further to find the value of the second variable.
-
Cross Multiplication Method:
Let’s consider the equations to be a1x+ b1y + c1=0 and a2x+ b2y+c2=0
Now as the name suggests we need to cross multiply to find the values of x and y. The formula to determine the answer will be –
x = b1c2- b1c1 a1b2- a2b1 and y = a2c1- a1c2 a1b2- a2b1
Solve the equation further using the formula to derive the answer.
Let’s revise briefly what we saw in the chapter before we proceed with example and practice questions, an equation in the form ax + by + c = 0, where a, b, c are real numbers, and a ≠ 0 and b ≠ 0, is called a linear equation with two variables. These equations have infinite solutions and represent a single line on the graph. These pair of lines can be represented graphically as intersecting lines, parallel lines, or coincident lines depending on the correlation between the coefficients. To solve these linear pair of equations with two variables we can use either of the three solution methods namely, substitution method or cross-multiplication method.
Let’s have a look at some sample questions and their answers and get a hand on some practice questions towards the end.
SAMPLE QUESTIONS
Question 1:
How many solutions does the pair of equations y = -8 and y = 0 have?
Solution:
y = -8 and y = 0 are parallel lines and thus have no solution.
Question 2:
If ax+by=a2-b2 and bx+ay=0, find the value of (x+y)
Solution:
ax+by=a2-b2
bx+ay=0
Adding … a(x+y) +b(x+y) =a2-b2
(x+y) (a+b) =a2-b2
x+y= a+b(a-b) (a+b)
Therefore, x+y=a-b
Question 3:
Solve the following pair of linear equations for x and y:
2x+3y=20; 7x+2y=53
Solution:
(2x+3y=20) x 2 …. Multiplying equation i by 2
(7x+2y=53) x -3 … Multiplying equation ii by -3
Adding the two equations 4x+6y= 40
-21x-6y=-159
-17x =-119
x=-119-17
x=7
Using x=7 in 2x+3y=20
2(7)+3y=20
14+3y=20
3y=20-14
y=63
y=2
Thus, the solution to the linear equations is x=7 and y=2.
Question 4:
Solve by elimination
3x=y+5; 5x-y=11
Solution:
3x-y= 5 ……… Subtracting equation two from equation one
5x-y=11
- + -
-2x =-6
x =3
Putting value of x in first equation, 3x-y=5 → 33-y=5→ 9-5=y → y=4
Thus, x=3; y=4
Question 5:
Find the two numbers whose sum is 85 and difference is 35
Solution:
Let’s consider the numbers to be x and y.
According to the question the two equations will be;
x+y=85 and x-y=±35
Solving both equations
x+y=85 x+y=85
x-y=35 x-y=-35
2x =120 2x = 50
x =60 x =25
Substituting these values of x in equation 1
60+y=85 25+y=85
y=85-60 y=85-25
y=25 y=60
Thus, the two numbers are 60 and 25.
Question 6:
A man earns ?600 per month more than his wife. One-tenth of the man’s salary and 1/6th of the wife’s salary amount to ?1,500, which is saved every month. Find their incomes
Solution:
Let’s consider wife’s monthly income to be = ? x
And man’s monthly income = ? (x+600)
As the question suggests,
110 x+600+ 16 x= ?1,500
3x+1,800+5x= ?45,000
8x=?45,000-?1,800
x=?3,43,2008=?5,400
Thus, wife’s income = 8x=?x=?5,400 and Man’s income = ?x+600=?6,000
PRACTICE QUESTIONS
- Solve the following pair of linear equations for x and y:
141x+93y=189;
93x+141y=45 - Solve by elimination:
3x-y=7
2x+5y+1=0 - Find the value of α and β for which the following pair of linear equations has infinite number of solutions:
2x+3y=7; ax+(α+β) y=28
- Solve for x and y: xa=yb; ax+by=a2 + b2
- The sum of the digits of a two-digit number is 8 and the difference between the number and that formed by reversing the digits is 18. Find the number.
- The age of the father is twice the sum of the ages of his 2 children. After 20 years, his age will be equal to the sum of the ages of his children. Find the age of the father.
- A two-digit number is seven times the sum of its digits. The number formed by reversing the digits is 18 less than the given number. Find the given number.
- Speed of a boat in still water is 15 km/h. It goes 30 km upstream and returns back at the same point in 4 hours 30 minutes. Find the speed of the stream.
- A man travels 300 km partly by train and partly by car. He takes 4 hours if the travels 60 km by train and the rest by car. If he travels 100 km by train and the remaining by car, he takes 10 minutes longer. Find the speeds of the train and the car separately.
Comments