NCERT Solutions · Class 9 Mathematics Two Variables, One Line

Substitute \(\displaystyle x=4,\ y=3\) in the left side. \[5x-6y = 5(4)-6(3) \] \[= 20-18 = 2 \] The left side equals the right side, so the pair satisfies the equation.Answer: Yes, \(\displaystyle (4,3)\) is a solution, since \(\displaystyle 5(4)-6(3)=2\).

Choose a value for one variable and solve for the other. (i) \[x=0:\ -3y=21 \Rightarrow y=-7 \] \[y=0:\ 7x=21 \Rightarrow x=3 \] \[7(0)-3(-7)=21, \qquad 7(3)-3(0)=21 \] (ii) \[x=1:\ 2+3y=5 \Rightarrow y=1 \] \[x=4:\ 8+3y=5 \Rightarrow y=-1 \] \[2(1)+3(1)=5, \qquad 2(4)+3(-1)=5 \] Any other pairs that satisfy the equations are equally valid.Answer: (i) \(\displaystyle (0,-7)\) and \(\displaystyle (3,0)\); (ii) \(\displaystyle (1,1)\) and \(\displaystyle (4,-1)\).

Substitute \(\displaystyle x=2,\ y=-1\) in each equation. \[2m(2)+3(-1)=7 \] \[4m-3=7 \Rightarrow m=\tfrac{5}{2} \] \[4(2)+n(-1)=-10 \] \[8-n=-10 \Rightarrow n=18 \]Answer: \(\displaystyle m=\tfrac{5}{2}\), \(\displaystyle n=18\).

Choose \(\displaystyle x\), find \(\displaystyle y\), and read the quadrant from the signs. (i) \[x=2:\ 10+3y=7 \Rightarrow y=-1 \] \[x=-1:\ -5+3y=7 \Rightarrow y=4 \] \(\displaystyle (2,-1)\) lies in quadrant IV; \(\displaystyle (-1,4)\) in quadrant II. (ii) \[x=2:\ 10-3y=7 \Rightarrow y=1 \] \[x=-1:\ -5-3y=7 \Rightarrow y=-4 \] \(\displaystyle (2,1)\) lies in quadrant I; \(\displaystyle (-1,-4)\) in quadrant III. (iii) \[x=1:\ -5+3y=7 \Rightarrow y=4 \] \[x=-2:\ 10+3y=7 \Rightarrow y=-1 \] \(\displaystyle (1,4)\) lies in quadrant I; \(\displaystyle (-2,-1)\) in quadrant III. (iv) \[x=-2:\ 10-3y=7 \Rightarrow y=1 \] \[x=1:\ -5-3y=7 \Rightarrow y=-4 \] \(\displaystyle (-2,1)\) lies in quadrant II; \(\displaystyle (1,-4)\) in quadrant IV. Any other two solutions in different quadrants are equally valid.Answer: (i) \(\displaystyle (2,-1)\) IV, \(\displaystyle (-1,4)\) II; (ii) \(\displaystyle (2,1)\) I, \(\displaystyle (-1,-4)\) III; (iii) \(\displaystyle (1,4)\) I, \(\displaystyle (-2,-1)\) III; (iv) \(\displaystyle (-2,1)\) II, \(\displaystyle (1,-4)\) IV.

The line in the given figure passes through \(\displaystyle A(7,0)\) and \(\displaystyle B(0,-3)\), both solutions: \[3(7)-7(0)=21, \qquad 3(0)-7(-3)=21 \] Now test \(\displaystyle C(2,3)\): \[3x-7y = 3(2)-7(3) = 6-21 = -15 \neq 21 \] On the line, \(\displaystyle x=2\) gives \[y=\frac{3(2)-21}{7}=-\frac{15}{7} \neq 3 \] So C is not on the line and does not satisfy the equation. No point off the line can satisfy it: the solutions of the equation are exactly the points of the line.Answer: No; C does not lie on the line, it does not satisfy the equation, and no point off the line satisfies it.

(i) False. Counterexample: \(\displaystyle x+y=2\) has the different solutions \[(0,2):\ 0+2=2, \qquad (1,1):\ 1+1=2 \] (ii) False. Counterexample: for \(\displaystyle x+y=1\) the origin gives \[0+0=0 \neq 1 \] So the line misses \(\displaystyle (0,0)\). A line \(\displaystyle ax+by+c=0\) passes through the origin only when \(\displaystyle c=0\). (iii) False. Counterexample: for \(\displaystyle 2x+3y=5\) \[2\left(\tfrac12\right)+3\left(\tfrac43\right)=1+4=5 \] so \(\displaystyle \left(\tfrac12,\tfrac43\right)\) is a rational solution. (iv) True. \[x=3 \iff 1\cdot x+0\cdot y-3=0 \] Here \(\displaystyle a=1,\ b=0\) are not both zero, so it is in standard form. (v) True. For every real \(\displaystyle u\), the pair \(\displaystyle \left(u,\ \frac{7-2u}{3}\right)\) is a solution: \[2u+3\cdot\frac{7-2u}{3}=7 \] (vi) False. \[2(1)+3(2)=8 \neq 7 \]Answer: (i) False; (ii) False; (iii) False; (iv) True; (v) True; (vi) False.

(i) The left sides differ by the factor $\displaystyle 2$: \[6x+8y-14 = 2\,(3x+4y-7) \] so one is zero exactly when the other is. Two solutions of the first also satisfy the second: \[(1,1):\ 3+4=7, \quad 6+8=14 \] \[\left(3,-\tfrac12\right):\ 9-2=7, \quad 18-4=14 \] (ii) Let \(\displaystyle (x_0,y_0)\) be any pair. \[ax_0+by_0=c \Rightarrow k(ax_0+by_0)=kc \Rightarrow kax_0+kby_0=kc \] Conversely, dividing by \(\displaystyle k \neq 0\): \[kax_0+kby_0=kc \Rightarrow ax_0+by_0=c \]Answer: Each equation is the other multiplied (or divided) by a non-zero number, so every solution of one satisfies the other; the solution sets are equal.