NCERT Class 10 Maths Exercise 3.1 Solution – Chapter 3

Solution:

1. $$\text{Let boys}=x,\text{girls}=y$$

   First, choose the variables. Let the number of boys be x and the number of girls be y.

2. $$\text{Equation 1: }x+y=10\cdots (1)$$

   There are 10 students in total, so x + y = 10. This is the first equation.

3. $$\text{Equation 2: }y-x=4\cdots (2)$$

   There are 4 more girls than boys, so y – x = 4. This is the second equation.

4. $$\overline{)\begin{array}{cccc}x& 0& 5& 10\\ y=10-x& 10& 5& 0\end{array}}$$

   Make a table of values for the first line, x + y = 10. When x = 0, y = 10. When x = 5, y = 5. When x = 10, y = 0. Plot these points on the graph and draw the line.

5. $$\overline{)\begin{array}{cccc}x& 0& 2& 5\\ y=x+4& 4& 6& 9\end{array}}$$

   Make a table for the second line, y – x = 4, which is y = x + 4. When x = 0, y = 4. When x = 2, y = 6. When x = 5, y = 9. Plot these points and draw the second line.

6. $$\text{Intersection point}=(3,7)$$

   The two lines meet at the point (3, 7). This is our solution.

7. $$\therefore \text{Boys}=x=3,\text{Girls}=y=7$$

   So the final answer is: there were 3 boys and 7 girls in the quiz. Check: 3 + 7 = 10 ✓ and 7 – 3 = 4 ✓

Solution:

1. $$\text{Let cost of 1 pencil}=x\text{ (Rs.)},\text{cost of 1 pen}=y\text{ (Rs.)}$$

   First, choose the variables. Let the cost of one pencil be Rs. x and the cost of one pen be Rs. y.

2. $$\begin{gathered} 5x+7y=50\cdots (1) \\ 7x+5y=46\cdots (2) \end{gathered}$$

   The two conditions give us 2 linear equations. The first is 5x + 7y = 50 and the second is 7x + 5y = 46.

3. $$\begin{array}{ccc}& x& y\\ 5x+7y=50& 10& 0\\ & 3& 5\\ & -4& 10\end{array}$$

   Now make a table of values for the first equation. When x = 10, y = 0. When x = 3, y = 5. When x = -4, y = 10.

4. $$\begin{array}{ccc}& x& y\\ 7x+5y=46& 8& -2\\ & 3& 5\\ & -2& 12\end{array}$$

   Now make a table for the second equation. When x = 8, y = -2. When x = 3, y = 5. When x = -2, y = 12.

5. $$\text{Lines intersect at point }(3,5)$$

   Plot both lines on the graph. They cross each other at the point (3, 5). This is our solution.

6. $$\text{Verification: }5(3)+7(5)=15+35=50\mathrm{✓},7(3)+5(5)=21+25=46\mathrm{✓}$$

   In the board exam, always check your answer. Put both values in both equations. The left side equals the right side, so the answer is correct.

7. $$x=3\Rightarrow \text{Cost of 1 pencil}=\text{Rs. }3,y=5\Rightarrow \text{Cost of 1 pen}=\text{Rs. }5$$

   So the final answer is: one pencil costs Rs. 3 and one pen costs Rs. 5.

Solution:

1. $$5x-4y+8=0\text{and}7x+6y-9=0$$

   We have 2 equations. To compare them, we will find the ratios.

2. $$a_1=5,b_1=-4,c_1=8\text{and}{a}_2=7,b_2=6,c_2=-9$$

   First, find the coefficients from both equations: a1 = 5, b1 = -4, c1 = 8, and a2 = 7, b2 = 6, c2 = -9.

3. $$\frac{a_1}{a_2}=\frac{5}{7}$$

   Now find the first ratio: a1/a2 = 5/7.

4. $$\frac{b_1}{b_2}=\frac{-4}{6}=\frac{-2}{3}$$

   The second ratio is b1/b2 = -4/6, which simplifies to -2/3.

5. $$\frac{a_1}{a_2}=\frac{5}{7}\mathrm{\ne }\frac{b_1}{b_2}=\frac{-2}{3}$$

   See, 5/7 and -2/3 are not equal. This is the condition for intersecting lines.

6. $$\frac{a_1}{a_2}\mathrm{\ne }\frac{b_1}{b_2}⟹\text{Lines intersect at a point}$$

   When a1/a2 ≠ b1/b2, the lines meet at one point. This means there is a unique solution.

Solution:

1. $$9x+3y+12=0\text{and}18x+6y+24=0$$

   We have 2 equations. The first is 9x + 3y + 12 = 0 and the second is 18x + 6y + 24 = 0.

2. $$a_1=9,b_1=3,c_1=12\text{and}{a}_2=18,b_2=6,c_2=24$$

   Now find the coefficients. From the first equation, a1 = 9, b1 = 3, c1 = 12. From the second equation, a2 = 18, b2 = 6, c2 = 24.

3. $$\frac{a_1}{a_2}=\frac{9}{18}=\frac{1}{2}$$

   Find the first ratio: a1/a2 = 9/18, which simplifies to 1/2.

4. $$\frac{b_1}{b_2}=\frac{3}{6}=\frac{1}{2}$$

   The second ratio is b1/b2 = 3/6, which also simplifies to 1/2.

5. $$\frac{c_1}{c_2}=\frac{12}{24}=\frac{1}{2}$$

   The third ratio is c1/c2 = 12/24, which also simplifies to 1/2.

6. $$\frac{a_1}{a_2}=\frac{b_1}{b_2}=\frac{c_1}{c_2}=\frac{1}{2}$$

   See, all three ratios are equal: a1/a2 = b1/b2 = c1/c2 = 1/2. This condition is very important!

7. $$\because \frac{a_1}{a_2}=\frac{b_1}{b_2}=\frac{c_1}{c_2}\Rightarrow \text{lines are coincident (infinitely many solutions)}$$

   When all three ratios are equal, the lines are coincident. This means both lines are really the same line, and they have infinitely many solutions.

Solution:

1. $$6x-3y+10=0\text{and}2x-y+9=0$$

   We have 2 equations. The first is 6x – 3y + 10 = 0 and the second is 2x – y + 9 = 0.

2. $$a_1=6,b_1=-3,c_1=10\text{and}{a}_2=2,b_2=-1,c_2=9$$

   Find the coefficients from both equations. From the first, a1 = 6, b1 = -3, c1 = 10. From the second, a2 = 2, b2 = -1, c2 = 9.

3. $$\frac{a_1}{a_2}=\frac{6}{2}=3$$

   Now find the first ratio: a1/a2 = 6/2 = 3.

4. $$\frac{b_1}{b_2}=\frac{-3}{-1}=3$$

   Find the second ratio: b1/b2 = -3/-1 = 3.

5. $$\frac{c_1}{c_2}=\frac{10}{9}$$

   Find the third ratio: c1/c2 = 10/9.

6. $$\frac{a_1}{a_2}=\frac{b_1}{b_2}=3\text{but}\frac{c_1}{c_2}=\frac{10}{9}$$

   See, a1/a2 = b1/b2 = 3, but c1/c2 = 10/9, which is not equal to 3.

7. $$\frac{a_1}{a_2}=\frac{b_1}{b_2}\mathrm{\ne }\frac{c_1}{c_2}\Rightarrow \text{Lines are Parallel}$$

   This is the condition for parallel lines! When a1/a2 = b1/b2 ≠ c1/c2, the two lines are parallel and have no solution. So this pair is inconsistent.

Solution:

1. $$3x+2y=5\Rightarrow 3x+2y-5=0$$

   The first equation is 3x + 2y = 5. Write it in standard form: 3x + 2y – 5 = 0.

2. $$2x-3y=7\Rightarrow 2x-3y-7=0$$

   The second equation is 2x – 3y = 7. Write it in standard form: 2x – 3y – 7 = 0.

3. $$a_1=3,b_1=2,c_1=-5\text{and}{a}_2=2,b_2=-3,c_2=-7$$

   Now find the coefficients from both equations: a1 = 3, b1 = 2, c1 = -5, and a2 = 2, b2 = -3, c2 = -7.

4. $$\frac{a_1}{a_2}=\frac{3}{2},\frac{b_1}{b_2}=\frac{2}{-3}=-\frac{2}{3},\frac{c_1}{c_2}=\frac{-5}{-7}=\frac{5}{7}$$

   Find all three ratios: a1/a2 = 3/2, b1/b2 = 2/-3 = -2/3, and c1/c2 = -5/-7 = 5/7.

5. $$\frac{a_1}{a_2}=\frac{3}{2}\mathrm{\ne }\frac{b_1}{b_2}=-\frac{2}{3}$$

   See, a1/a2 = 3/2 and b1/b2 = -2/3. These two ratios are not equal.

6. $$\frac{a_1}{a_2}\mathrm{\ne }\frac{b_1}{b_2}\Rightarrow \text{Lines intersect}\Rightarrow \text{Unique solution}$$

   When a1/a2 ≠ b1/b2, the lines intersect. This means there is a unique solution.

7. $$\therefore \text{The pair of equations is CONSISTENT}$$

   So this pair of linear equations is consistent. It has one unique solution.

Solution:

1. $$2x-3y=8\Rightarrow a_1=2,b_1=-3,c_1=-8$$

   The first equation is 2x – 3y = 8. From it we get a1 = 2, b1 = -3, and c1 = -8.

2. $$4x-6y=9\Rightarrow a_2=4,b_2=-6,c_2=-9$$

   The second equation is 4x – 6y = 9. So a2 = 4, b2 = -6, and c2 = -9.

3. $$\frac{a_1}{a_2}=\frac{2}{4}=\frac{1}{2},\frac{b_1}{b_2}=\frac{-3}{-6}=\frac{1}{2},\frac{c_1}{c_2}=\frac{-8}{-9}=\frac{8}{9}$$

   Now find all three ratios: a1/a2 = 2/4 = 1/2, b1/b2 = -3/-6 = 1/2, and c1/c2 = -8/-9 = 8/9.

4. $$\frac{a_1}{a_2}=\frac{b_1}{b_2}=\frac{1}{2}\text{but}\frac{c_1}{c_2}=\frac{8}{9}\mathrm{\ne }\frac{1}{2}$$

   See, a1/a2 and b1/b2 are both equal to 1/2, but c1/c2 = 8/9, which is not equal to 1/2.

5. $$\frac{a_1}{a_2}=\frac{b_1}{b_2}\mathrm{\ne }\frac{c_1}{c_2}\Rightarrow \text{Parallel lines, No solution}$$

   When a1/a2 = b1/b2 ≠ c1/c2, the lines are parallel and there is no solution.

6. $$\therefore \text{The pair of equations is }\mathbf{\text{Inconsistent}}$$

   So this pair of linear equations is inconsistent. It has no solution.

Solution:

1. $$\frac{3}{2}x+\frac{5}{3}y=7\Rightarrow 9x+10y=42$$

   First, simplify the equation. Multiply both sides by 6 to get 9x + 10y = 42.

2. $$a_1=9,b_1=10,c_1=-42(9x+10y-42=0)$$

   Now write it in standard form. The coefficients of the first equation are a1 = 9, b1 = 10, and c1 = -42.

3. $$a_2=9,b_2=-10,c_2=-14(9x-10y-14=0)$$

   From the second equation, 9x – 10y = 14, the coefficients are a2 = 9, b2 = -10, and c2 = -14.

4. $$\frac{a_1}{a_2}=\frac{9}{9}=1$$

   Now find the ratios: a1/a2 = 9/9 = 1.

5. $$\frac{b_1}{b_2}=\frac{10}{-10}=-1$$

   b1/b2 = 10/-10 = -1.

6. $$\frac{a_1}{a_2}=1\mathrm{\ne }-1=\frac{b_1}{b_2}$$

   See, a1/a2 = 1 but b1/b2 = -1. The two ratios are not equal.

7. $$\frac{a_1}{a_2}\mathrm{\ne }\frac{b_1}{b_2}⟹\text{Consistent (unique solution)}$$

   When a1/a2 ≠ b1/b2, the pair of equations is consistent and has exactly 1 unique solution.

Solution:

1. $$5x-3y=11\text{and}-10x+6y=-22$$

   We have 2 equations. Let us write them in the standard form a1x + b1y + c1 = 0.

2. $$a_1=5,b_1=-3,c_1=-11\text{and}{a}_2=-10,b_2=6,c_2=22$$

   From the first equation take a1 = 5, b1 = -3, c1 = -11. From the second take a2 = -10, b2 = 6, c2 = 22.

3. $$\frac{a_1}{a_2}=\frac{5}{-10}=-\frac{1}{2}$$

   Now find the ratios: a1/a2 = 5/-10 = -1/2.

4. $$\frac{b_1}{b_2}=\frac{-3}{6}=-\frac{1}{2}$$

   b1/b2 = -3/6 = -1/2.

5. $$\frac{c_1}{c_2}=\frac{-11}{22}=-\frac{1}{2}$$

   c1/c2 = -11/22 = -1/2.

6. $$\frac{a_1}{a_2}=\frac{b_1}{b_2}=\frac{c_1}{c_2}=-\frac{1}{2}$$

   All three ratios are equal: a1/a2 = b1/b2 = c1/c2. This condition tells us that the lines are coincident and have infinitely many solutions.

7. $$\therefore \text{Pair is Consistent (Infinitely Many Solutions)}$$

   So this pair of linear equations is consistent and has infinitely many solutions. This is what you should write in the board exam.

Solution:

1. $$\frac{4}{3}x+2y=8\Rightarrow a_1=\frac{4}{3},b_1=2,c_1=8$$

   Find the values from the first equation: a1 = 4/3, b1 = 2, and c1 = 8.

2. $$2x+3y=12\Rightarrow a_2=2,b_2=3,c_2=12$$

   Take the values from the second equation: a2 = 2, b2 = 3, and c2 = 12.

3. $$\frac{a_1}{a_2}=\frac{\frac{4}{3}}{2}=\frac{4}{3\times 2}=\frac{4}{6}=\frac{2}{3}$$

   Now find the ratios: a1/a2 = (4/3) ÷ 2, which simplifies to 2/3.

4. $$\frac{b_1}{b_2}=\frac{2}{3}$$

   b1/b2 = 2/3. See, this is also 2/3.

5. $$\frac{c_1}{c_2}=\frac{8}{12}=\frac{2}{3}$$

   c1/c2 = 8/12, which simplifies to 2/3.

6. $$\frac{a_1}{a_2}=\frac{b_1}{b_2}=\frac{c_1}{c_2}=\frac{2}{3}$$

   All three ratios are equal: 2/3 = 2/3 = 2/3. This condition tells us that the lines are coincident.

7. $$\therefore \text{Lines are coincident}\Rightarrow \text{Infinitely many solutions}\Rightarrow \text{CONSISTENT}$$

   When all three ratios are equal, the lines are coincident, there are infinitely many solutions, and the pair of equations is consistent.

Solution:

1. $$\text{Line 1: }x+y=5\Rightarrow \frac{a_1}{a_2}=\frac{1}{2},\frac{b_1}{b_2}=\frac{1}{2},\frac{c_1}{c_2}=\frac{5}{10}=\frac{1}{2}$$

   First, we check the ratios: a1/a2 = 1/2, b1/b2 = 1/2, and c1/c2 = 5/10 = 1/2.

2. $$\frac{a_1}{a_2}=\frac{b_1}{b_2}=\frac{c_1}{c_2}=\frac{1}{2}\Rightarrow \text{Lines are COINCIDENT (Consistent)}$$

   All three ratios are equal, so the two lines are coincident. On the graph they lie on top of each other, and they have infinitely many solutions. This pair is consistent.

3. $$\begin{array}{cccc}x& 0& 2& 5\\ y=5-x& 5& 3& 0\end{array}\text{(Line 1: }x+y=5\text{)}$$

   Now make a table for Line 1, x + y = 5. When x = 0, y = 5. When x = 2, y = 3. When x = 5, y = 0.

4. $$\begin{array}{cccc}x& 0& 2& 5\\ y=\frac{10-2x}{2}& 5& 3& 0\end{array}\text{(Line 2: }2x+2y=10\text{)}$$

   Simplify Line 2, 2x + 2y = 10, and you get x + y = 5. It is the same equation! Its table gives exactly the same points.

5. $$\text{Both lines pass through }(0,5),(2,3),(5,0)\text{ — they COINCIDE on the graph}$$

   Look at the graph. Both lines lie exactly on top of each other, because both equations are the same.

6. $$\text{Solution: Infinitely many solutions, e.g. }(0,5),(2,3),(5,0),\dots$$

   So this pair has infinitely many solutions. Every point on the line is a solution. The pair is consistent.

Solution:

1. $$\text{Line 1: }x-y=8\Rightarrow y=x-8$$

   The first equation is x – y = 8. Write it in the form y = x – 8.

2. $$\begin{array}{cc}x& y=x-8\\ 8& 0\\ 10& 2\\ 6& -2\end{array}$$

   Now find 3 points for Line 1. When x = 8, y = 0. When x = 10, y = 2. When x = 6, y = -2.

   Related Posts

   • NCERT Class 10 Maths Exercise 1.1 Solution – Real Numbers

     Free NCERT Class 10 Maths Exercise 1.1 Solutions for Real Numbers. Step-by-step answers for prime…

   • NCERT Class 10 Maths Exercise 2.1 Solution – Polynomials

     NCERT Class 10 Maths Exercise 2.1 Solutions – Polynomials. Find number of zeroes from graphs…

   • NCERT Class 10 Maths Exercise 1.2 Solution – Real Numbers

     NCERT Class 10 Maths Exercise 1.2 Solution – Step-by-step proofs that sqrt(5), 3+2sqrt(5), 1/sqrt(2), 7sqrt(5)…

3. $$\text{Line 2: }3x-3y=16\Rightarrow y=x-\frac{16}{3}$$

   The second equation is 3x – 3y = 16. Simplifying it gives y = x – 16/3.

4. $$\begin{array}{cc}x& y=x-\frac{16}{3}\\ \frac{16}{3}& 0\\ \frac{22}{3}& 2\\ \frac{10}{3}& -2\end{array}$$

   Now find points for Line 2. When x = 16/3, y = 0. When x = 22/3, y = 2. When x = 10/3, y = -2.

5. $$\frac{a_1}{a_2}=\frac{1}{3},\frac{b_1}{b_2}=\frac{-1}{-3}=\frac{1}{3},\frac{c_1}{c_2}=\frac{8}{\frac{16}{3}}=\frac{3}{2}$$

   Now check consistency: a1/a2 = 1/3 and b1/b2 = 1/3, but c1/c2 = 3/2. This one is different!

6. $$\frac{a_1}{a_2}=\frac{b_1}{b_2}\mathrm{\ne }\frac{c_1}{c_2}\Rightarrow \text{Lines are Parallel}$$

   When a1/a2 = b1/b2 ≠ c1/c2, the lines are parallel and the equations are inconsistent. There is no solution.

7. $$\text{Both lines have slope }=1,\text{ but different y-intercepts: }-8\text{ and }-\frac{16}{3}$$

   Look at the graph. Both lines have slope 1, so they are parallel and never meet.

8. $$\therefore \text{Inconsistent}\Rightarrow \text{No Solution (Parallel Lines)}$$

   So this pair of linear equations is inconsistent and has no solution.

Solution:

1. $$2x+y-6=0\Rightarrow y=6-2x$$

   Solve the first equation for y: y = 6 – 2x.

2. $$\begin{array}{cc}x& y=6-2x\\ 0& 6\\ 1& 4\\ 3& 0\end{array}$$

   Make 3 points for the first line. Put x = 0, 1, and 3 and find the values of y.

3. $$4x-2y-4=0\Rightarrow y=2x-2$$

   Solve the second equation for y: y = 2x – 2.

4. $$\begin{array}{cc}x& y=2x-2\\ 0& -2\\ 1& 0\\ 3& 4\end{array}$$

   Make 3 points for the second line too. Put x = 0, 1, and 3 and find the values of y.

5. $$\frac{a_1}{a_2}=\frac{2}{4}=\frac{1}{2},\frac{b_1}{b_2}=\frac{1}{-2}=-\frac{1}{2}$$

   Check consistency: a1/a2 = 1/2, but b1/b2 = -1/2. The two ratios are different, so the lines will intersect. This means the system is consistent!

6. $$\frac{a_1}{a_2}\mathrm{\ne }\frac{b_1}{b_2}\Rightarrow \text{Consistent (unique solution)}$$

   Since a1/a2 ≠ b1/b2, the two lines meet at one point. So there is one unique solution.

7. $$2x+(2x-2)-6=0\Rightarrow 4x=8\Rightarrow x=2$$

   To find the meeting point, put y = 2x – 2 in the first equation. This gives 4x = 8, so x = 2.

8. $$x=2,y=2(2)-2=2\Rightarrow \text{Solution: }(2,2)$$

   Put x = 2 to find y, which gives y = 2. So the two lines meet at the point (2, 2) on the graph. This is our answer!

Solution:

1. $$\text{Eq 1: }2x-2y-2=0\Rightarrow y=x-1$$

   The first equation is 2x – 2y – 2 = 0. Simplify it by dividing both sides by 2. This gives y = x – 1.

2. $$\begin{array}{cc}x& y=x-1\\ 0& -1\\ 1& 0\\ 3& 2\end{array}$$

   Now make a table of values for Line 1. When x = 0, y = -1. When x = 1, y = 0. When x = 3, y = 2.

3. $$\text{Eq 2: }4x-4y-5=0\Rightarrow y=x-\frac{5}{4}$$

   The second equation is 4x – 4y – 5 = 0. Simplify it by dividing both sides by 4. This gives y = x – 5/4.

4. $$\begin{array}{cc}x& y=x-\frac{5}{4}\\ 0& -1.25\\ 1& -0.25\\ 3& 1.75\end{array}$$

   Make a table of values for Line 2. When x = 0, y = -1.25. When x = 1, y = -0.25. When x = 3, y = 1.75.

5. $$\frac{a_1}{a_2}=\frac{2}{4}=\frac{1}{2},\frac{b_1}{b_2}=\frac{-2}{-4}=\frac{1}{2},\frac{c_1}{c_2}=\frac{-2}{-5}=\frac{2}{5}$$

   Now check consistency: a1/a2 = 1/2 and b1/b2 = 1/2, but c1/c2 = 2/5, which is different.

6. $$\frac{a_1}{a_2}=\frac{b_1}{b_2}\mathrm{\ne }\frac{c_1}{c_2}\Rightarrow \text{Parallel lines, No solution}$$

   This is the condition a1/a2 = b1/b2 ≠ c1/c2. It means the two lines are parallel and there is no solution, so the pair is INCONSISTENT.

7. $$\text{Lines are PARALLEL}\Rightarrow \text{INCONSISTENT}\Rightarrow \text{No Solution}$$

   Look at the graph. The two lines do not meet anywhere. So this pair of linear equations is INCONSISTENT and has no solution.

Solution:

1. $$\text{Garden fencing cost}\propto \text{Perimeter}=2(l+b)$$

   In real life, when we put a fence around a rectangular garden, the cost depends on the perimeter. That is why it is useful to find the length and width.

2. $$\text{Let width}=x\text{ m, length}=y\text{ m}$$

   First, choose the variables. Let the width of the garden be x metres and the length be y metres.

3. $$y=x+4\Rightarrow y-x=4\cdots (1)$$

   We are told that the length is 4 metres more than the width. So our first equation is y – x = 4.

4. $$\text{Half perimeter}=\frac{1}{2}\times 2(x+y)=x+y=36\cdots (2)$$

   The perimeter of a rectangle is 2(x + y). So half the perimeter is x + y = 36. This is our second equation.

5. $$\text{Add (1) and (2):}(y-x)+(y+x)=4+36$$

   Now add both equations. On the left side, x and -x cancel out.

6. $$2y=40\Rightarrow y=20$$

   2y = 40, so y = 20. This means the length is 20 metres.

7. $$x+y=36\Rightarrow x+20=36\Rightarrow x=16$$

   Now put the value of y in equation 2. x + 20 = 36, so x = 16. This means the width is 16 metres.

8. $$\text{Verify: }y-x=20-16=4\mathrm{✓}x+y=16+20=36\mathrm{✓}$$

   Check both conditions: length – width = 4 and x + y = 36. Both are correct!

9. $$\text{Length}=20\text{ m},\text{Width}=16\text{ m}$$

   So the final answer is: the length of the garden is 20 metres and the width is 16 metres.

Solution:

1. $$\text{Given: }2x+3y-8=0$$

   We are given a linear equation: 2x + 3y – 8 = 0. We must write a second equation so that the two lines intersect.

2. $$\text{Condition for intersecting lines: }\frac{a_1}{a_2}\mathrm{\ne }\frac{b_1}{b_2}$$

   For intersecting lines, the condition is a1/a2 ≠ b1/b2. This means the ratios of the coefficients of the two equations must be different.

3. $$a_1=2,b_1=3,c_1=-8$$

   In the given equation, a1 = 2, b1 = 3, and c1 = -8.

4. $$\text{Choose: }{a}_2=3,b_2=2,c_2=-7$$

   For the second equation, we choose a2 = 3 and b2 = 2, so that the ratios are different.

5. $$\frac{a_1}{a_2}=\frac{2}{3},\frac{b_1}{b_2}=\frac{3}{2},\frac{2}{3}\mathrm{\ne }\frac{3}{2}\mathrm{✓}$$

   Let us check: a1/a2 = 2/3 and b1/b2 = 3/2. The two ratios are different, so the condition is met.

6. $$\therefore 3x+2y-7=0$$

   So our second linear equation is 3x + 2y – 7 = 0. Together with 2x + 3y – 8 = 0, it forms intersecting lines.

Solution:

1. $$\text{Given: }2x+3y-8=0$$

   We are given a linear equation: 2x + 3y – 8 = 0. We must make a second equation whose line is parallel to it.

2. $$\text{Parallel condition: }\frac{a_1}{a_2}=\frac{b_1}{b_2}\mathrm{\ne }\frac{c_1}{c_2}$$

   For parallel lines, the condition is a1/a2 = b1/b2, but this must not be equal to c1/c2. This is the most important rule.

3. $$a_1=2,b_1=3,c_1=-8$$

   From the given equation, our coefficients are a1 = 2, b1 = 3, and c1 = -8.

4. $$\text{Choose }{a}_2=4,b_2=6,c_2=-24\text{ ?}$$

   If we take a2 = 4, b2 = 6, c2 = -24, then 2/4 = 3/6 = 8/24. That would be the same line, not a parallel line! Do not make this mistake.

5. $$\text{Choose }{a}_2=4,b_2=6,c_2=-9$$

   So we keep c2 different. We take a2 = 4, b2 = 6, and c2 = -9.

6. $$\frac{2}{4}=\frac{3}{6}=\frac{1}{2},\frac{-8}{-9}=\frac{8}{9}$$

   Let us check: 2/4 = 3/6 = 1/2, so these are equal. But 8/9 is not equal to 1/2. So the parallel condition is met!

7. $$\therefore \text{New equation: }4x+6y-9=0$$

   So our new equation is 4x + 6y – 9 = 0. Together with 2x + 3y – 8 = 0, it forms parallel lines.

Solution:

1. $$\text{Given: }2x+3y-8=0$$

   We are given a linear equation: 2x + 3y – 8 = 0. We must write an equation that forms coincident lines with it.

2. $$\text{Coincident lines}\Rightarrow \frac{a_1}{a_2}=\frac{b_1}{b_2}=\frac{c_1}{c_2}$$

   For coincident lines, the condition is a1/a2 = b1/b2 = c1/c2. This means the two equations must be proportional.

3. $$a_1=2,b_1=3,c_1=-8$$

   Find the coefficients from the given equation: a1 = 2, b1 = 3, and c1 = -8.

4. $$\text{Multiply entire equation by }k=2$$

   Now multiply the whole equation by any non-zero constant. We take k = 2.

5. $$2(2x+3y-8)=0$$

   Multiply every term inside the bracket by 2: 2 × 2x, 2 × 3y, and 2 × (-8).

6. $$\frac{2}{4}=\frac{3}{6}=\frac{-8}{-16}=\frac{1}{2}\mathrm{✓}$$

   Check: a1/a2 = 2/4, b1/b2 = 3/6, and c1/c2 = -8/-16. All three are equal, so the condition is met!

7. $$4x+6y-16=0$$

   So our new equation is 4x + 6y – 16 = 0. Together with the given equation, it forms coincident lines.

Solution:

1. $$\text{Line 1: }x-y+1=0\Rightarrow y=x+1$$

   The first equation is x – y + 1 = 0. Rearrange it to get y = x + 1. Now make a table of values for it.

2. $$\overline{)\begin{array}{cccc}x& 0& 1& 2\\ y=x+1& 1& 2& 3\end{array}}$$

   Put x = 0 to get y = 1. Put x = 1 to get y = 2. Put x = 2 to get y = 3. The three points are (0, 1), (1, 2), and (2, 3).

3. $$\text{Line 2: }3x+2y-12=0\Rightarrow y=\frac{12-3x}{2}$$

   The second equation is 3x + 2y – 12 = 0. Rearranging gives y = (12 – 3x)/2. Now make its table too.

4. $$\overline{)\begin{array}{cccc}x& 0& 2& 4\\ y=\frac{12-3x}{2}& 6& 3& 0\end{array}}$$

   Put x = 0 to get y = 6. Put x = 2 to get y = 3. Put x = 4 to get y = 0. The points are (0, 6), (2, 3), and (4, 0).

5. $$x-y+1=0\text{ and }3x+2y-12=0\text{ solve}$$

   Now find the point where the two lines meet. From the first equation, y = x + 1. Put this in the second equation.

6. $$3x+2(x+1)-12=0\Rightarrow 5x=10\Rightarrow x=2,y=3$$

   3x + 2(x + 1) – 12 = 0, so 5x = 10. This gives x = 2 and y = x + 1 = 3. The two lines meet at the point (2, 3).

7. $$\text{On the x-axis: }y=0\Rightarrow (-1,0)\text{ and }(4,0)$$

   On the x-axis, put y = 0. In Line 1, y = 0 gives x = -1. In Line 2, y = 0 gives x = 4. So the two vertices on the x-axis are (-1, 0) and (4, 0).

8. $$\text{Vertices: }(-1,0),(4,0),(2,3)$$

   So the three vertices of the triangle are (-1, 0), (4, 0), and (2, 3). This is our final answer.