‘Linear Equations’ is an important topic for RRB exams. The standard form of a two-variable linear equation is ax+by+c = 0 where x & y are two variables and a & b are not equal to zero. Only one linear equation of two variables cannot be solved without extra information or other equations. There are mainly four methods (Elimination, Substitution, Cross multiplication, and Determinant) of solving a two-variable based linear equation. From these four methods, mostly two methods (Elimination and Substitution) are used for solving which are given below with the help of examples.
Q) If 7p+8q = 52 and 5p-3q = 11, then find out the value of p.
Sol. 7p+8q = 52 Eq.(i)
5p-3q = 11 Eq.(ii)
Here we will use elimination method to get the values of p and q.
Multiply Eq.(i) by 3 and Eq.(ii) by 8.
21p+24q = 156 Eq.(iii)
40p-24q = 88 Eq.(iv)
Add Eq.(iii) and Eq.(iv).
61p = 156+88 = 244
p = 4
Q) If 11p-5q = 12 and 7p+4q = 22, then find out the value of (p-q).
Sol. 11p-5q = 12 Eq.(i)
7p+4q = 22
4q = 22-7p
q = 5.5-1.75p Eq.(ii)
Put Eq.(ii) in Eq.(i).
11p-5(5.5-1.75p) = 12
11p-27.5+8.75p = 12
19.75p = 12+27.5
19.75p = 39.5
p = 2
Put the value of ‘p’ in Eq.(i).
$$11\times2-5q = 12$$
22-5q = 12
5q = 22-12
q = 10
q = 2
value of (p-q) = 2-2 = 0
