Chapter 4 Linear Equations in Two Variables Ex 4.2 Solutions
Question - 1 : - Which one of the following options is true, and why?
y = 3x + 5 has
(i) a unique solution, (ii) only two solutions, (iii) infinitely many solutions
Answer - 1 : -
y = 3x + 5 is a linear equation in two variables and it has infinite possible solutions. As for every value of x, there will be a value of y satisfying the above equation and vice-versa.
Hence, the correct answer is (iii).
Question - 2 : - Write four solutions for each of the following equations:
(i) 2x + y = 7 (ii) ╧Аx + y = 9 (iii) x = 4y
Answer - 2 : -
(i) 2x┬а+┬аy┬а= 7
For┬аx┬а= 0,
2(0) +┬аy┬а= 7
тЗТ┬аy┬а= 7
Therefore, (0, 7) is a solution of this equation.
For┬аx┬а= 1,
2(1) +┬аy┬а= 7
тЗТ┬аy┬а= 5
Therefore, (1, 5) is a solution of this equation.
For┬аx┬а= тИТ1,
2(тИТ1) +┬аy┬а= 7
тЗТ┬аy┬а= 9
Therefore, (тИТ1, 9) is a solution of this equation.
For┬аx┬а= 2,
2(2) +┬аy┬а= 7
тЗТy┬а= 3
Therefore, (2, 3) is a solution of this equation.
(ii) ╧Аx┬а+┬аy┬а= 9
For┬аx┬а= 0,
╧А(0) +┬аy┬а= 9
тЗТ┬аy┬а= 9
Therefore, (0, 9) is a solution of this equation.
For┬аx┬а= 1,
╧А(1) +┬аy┬а= 9
тЗТy┬а= 9 тИТ╧А
Therefore, (1, 9 тИТ ╧А) is a solution of this equation.
For┬аx┬а= 2,
╧А(2) +┬аy┬а= 9
тЗТ┬аy┬а= 9 тИТ 2╧А
Therefore, (2, 9 тИТ2╧А) is a solution of this equation.
For┬аx┬а= тИТ1,
╧А(тИТ1) +┬аy┬а= 9
тЗТ┬аy┬а= 9 + ╧А
тЗТ (тИТ1, 9 + ╧А) is a solution of this equation.
(iii)┬аx┬а= 4y
For┬аx┬а= 0,
0 = 4y
тЗТ┬аy┬а= 0
Therefore, (0, 0) is a solution of this equation.
For┬аy┬а= 1,
x┬а= 4(1) = 4
Therefore, (4, 1) is a solution of this equation.
For y = тИТ1,
x┬а= 4(тИТ1)
тЗТ┬аx┬а= тИТ4
Therefore, (тИТ4, тИТ1) is a solution of this equation.
For┬аx┬а= 2,
2 = 4y
тЗТ
Therefore,┬а
is a solution of this equation.
Question - 3 : - Check which of the following are solutions of the equation┬аx┬атИТ2y┬а= 4 and which are not:
(i) (0, 2 (ii) (2, 0) (iii) (4, 0)
(iv)┬а
┬а(v) (1, 1)
Answer - 3 : -
(i) (0, 2)
Putting┬аx┬а= 0 and┬аy┬а= 2 in the L.H.Sof the given equation,
x┬атИТ 2y┬а= 0 тИТ2├Ч2 = тИТ 4 тЙа 4
L.H.S┬атЙа┬аR.H.S
Therefore, (0, 2) is not a solution of this equation.
(ii) (2, 0)
Putting┬аx┬а= 2 and┬аy┬а= 0 in the L.H.Sof the given equation,
x┬атИТ 2y┬а= 2 тИТ2 ├Ч 0 = 2 тЙа 4
L.H.S┬атЙа┬аR.H.S
Therefore, (2, 0) is not a solution of this equation.
(iii) (4, 0)
Putting┬аx┬а= 4 and┬аy┬а= 0 in the L.H.Sof the given equation,
x┬атИТ 2y┬а= 4 тИТ2(0)
= 4 = R.H.S
Therefore, (4, 0) is a solution of this equation.
(iv)Putting┬а
┬аand┬а
in the L.H.S of the given equation, L.H.S┬атЙа┬аR.H.S
Therefore,┬аis not a solution of this equation.
(v) (1, 1)
Putting┬аx┬а= 1 and┬аy┬а= 1 in the L.H.Sof the given equation,
x┬атИТ 2y┬а= 1 тИТ2(1) = 1 тИТ 2 = тИТ 1 тЙа 4
L.H.S┬атЙа┬аR.H.S
Therefore, (1, 1) is not a solution of this equation.
Question - 4 : - Find the value of k, if x = 2, y = 1 is a solution of the equation 2x + 3y = k.
Answer - 4 : -
Putting x = 2 and y = 1 in the given equation,
2x + 3y = k
тЗТ 2(2) + 3(1) = k
тЗТ 4 + 3 = k
тЗТ k = 7
Therefore, the value of k is 7.