Pair of Linear Equations in Two Variable
Two linear equations that have the same two variables are known as a pair of linear equations in two variables. Learn about Linear equations fromÂ AP SSC Class 10 Maths Chapter 4 Pair of Linear Equations in Two Variable.
\(a_{1}x+b_{1}x+c_{1}=0\,(a_{1}^2+b_{1}^2\neq 0)\) \(a_{2}x+b_{2}x+c_{2}=0\,(a_{2}^2+b_{2}^2\neq 0)\)where \(a_1,a_2,b_1,b_2,c_1,c_2\) are all real numbers.
We use the following methods to find solutions to a pair of linear equations:
 Model Method
 Graphical Method
 Algebraic methods – Substitution method and Elimination method
There exists a relation between the coefficients and nature of the system of equations. Following are the relationship:
 \(\frac{a_1}{a_2}\neq \frac{b_1}{b_2}\), then the pair of linear equations is consistent
 \(\frac{a_1}{a_2}=\frac{b_1}{b_2}\neq \frac{c_1}{c_2}\), then the pair of linear equations is inconsistent.
 \(\frac{a_1}{a_2}=\frac{b_1}{b_2}=\frac{c_1}{c_2}\), then the pair of linear equations is dependent and consistent.
Let us look at a few solved questions from the chapter to better understand a pair of linear equations.
Class 10 Maths Chapter 4 Pair of Linear Equations in Two Variable Solutions

 Solve the following pair of equations by reducing them to a pair of linear equations.
Solution:
Let us consider the following
\(\frac{1}{x1}=u\) â€¦â€¦â€¦.(1) \(\frac{1}{y2}=v\)………….(2)Hence, the equations becomes
\(5u+v=2\)…………..(3) \(6u+3v=1\)…………(4)From (1),
\(v=25u\)Substituting the above in (4), we get
\(6u3(25u)=1\)Solving,
\(6u6+15u=1\) \(21u6=1\) \(21u=7\) \(u=\frac{7}{21}=\frac{1}{3}\)To find the value of v, substitute the above value of u in equation (3)
\(5(\frac{1}{3})+v=2\) \(\frac{5}{3}+v=2\) \(v=2\frac{5}{3}\) \(v=\frac{65}{3}=\frac{1}{3}\)Hence,
\(v=\frac{1}{3}\)To find the value of x and y, substitute the values of u and v in (1) and (2),
\(u=\frac{1}{x1}\) \(\frac{1}{3}=\frac{1}{x1}\) \(x1=3\) \(x=4\) \(v=\frac{1}{y2}\) \(\frac{1}{3}=\frac{1}{y2}\) \({y2}=3\) \(y=5\)The value of x and y are 4 and 5 respectively for the given pair of equations.
2.Â A publisher is planning to produce a new textbook. The fixed costs (reviewing, editing, typesetting
and so on) are Rs, 31.25 per book. Besides that, he also spends another Rs, 320000 in producing the book. The wholesale price (the amount received by the publisher) is Rs, 43.75 per book. How many books must the publisher sell to break even, i.e., so that the cost will equal revenues?
Solutions:Â The publisher breaks even when costs are equal to the revenues. Given that “x” represents the number of books printed and sold and “y” is the breakeven point, then the cost and revenue equations for the publisher are
The Cost equation given by “y” = 320000 + 31.25x (1)
The Revenue equation also given by “y” = 43.75x (2)
Using the second equation and substituting for “y” in the first equation, we get
43.75x = 3,20,000 + 31.25x
12.5x = 3,20,000
x = 3, 20,000
12.5 = 25,600
Thus, the publisher will break even when 25,600 books are printed and sold.
Stay tuned to BYJU’S to get the latest notification on SSC exam along with AP SSC model papers, exam pattern, marking scheme and more.