EXERCISE 3.1
Pair Of Linear Equations In Two Variables • 7 Questions
Question 1
Hint available
Form the pair of linear equations in the following problems, and find their solutions graphically. (i) 10 students of Class X took part in a quiz. If the number of girls is 4 more than the number of boys, find the number of boys and girls who took part in the quiz. Fig. 3.2 29 (ii) 5 pencils and 7 pens together cost ` 50, whereas 7 pencils and 5 pens together cost ` 46. Find the cost of one pencil and that of one pen.
Key Idea
Translate the word statements into two linear equations in two variables, then locate the point of intersection of the two straight lines (graphical solution). The coordinates of the intersection give the required numbers or costs.
Step-by-Step Solution
### Part (i)
1. Let \(b\) = number of boys and \(g\) = number of girls.
2. Total students: \(b+g = 10\) (Equation 1)
3. Girls are 4 more than boys: \(g = b + 4\) (Equation 2)
4. Graphical method – Plot the two lines on the \(b\)-\(g\) plane:
- For Eq. 1, take \(b=0\) ⇒ \(g=10\) and \(g=0\) ⇒ \(b=10\). Draw the line joining (0,10) and (10,0).
- For Eq. 2, take \(b=0\) ⇒ \(g=4\) and \(b=6\) ⇒ \(g=10\). Draw the line joining (0,4) and (6,10).
5. The two lines intersect at the point where both equations are satisfied. Solving algebraically gives:
\[g = b+4 \Rightarrow b+(b+4)=10 \Rightarrow 2b=6 \Rightarrow b=3\]
Substituting back: \(g = 3+4 = 7\).
6. Hence the intersection point is \((b,g) = (3,7)\).
Answer: 3 boys and 7 girls.
### Part (ii)
1. Let \(p\) = cost of one pencil (in rupees) and \(q\) = cost of one pen.
2. From the data:
- 5 pencils + 7 pens cost \(\text{Rs }50\): \(5p + 7q = 50\) (Equation 1)
- 7 pencils + 5 pens cost \(\text{Rs }46\): \(7p + 5q = 46\) (Equation 2)
3. Graphical method – Plot the two lines on the \(p\)-\(q\) plane:
- For Eq. 1, choose \(p=0\) ⇒ \(q=\frac{50}{7}\approx7.14\) and \(q=0\) ⇒ \(p=10\). Plot points (0,7.14) and (10,0).
- For Eq. 2, choose \(p=0\) ⇒ \(q=\frac{46}{5}=9.2\) and \(q=0\) ⇒ \(p=\frac{46}{7}\approx6.57\). Plot points (0,9.2) and (6.57,0).
4. The intersection of the two lines gives the simultaneous solution. Solving algebraically (as done for checking):
\[\begin{aligned}
5p+7q &= 50 \quad\text{(1)}\\
7p+5q &= 46 \quad\text{(2)}
\end{aligned}\]
Multiply (1) by 5 and (2) by 7:
\[25p+35q = 250\]
\[49p+35q = 322\]
Subtract: \(24p = 72 \Rightarrow p = 3\).
Substitute \(p=3\) into (1): \(5(3)+7q = 50 \Rightarrow 15+7q = 50 \Rightarrow 7q = 35 \Rightarrow q = 5\).
5. Hence the intersection point is \((p,q) = (3,5)\).
Answer: Cost of one pencil = Rs 3, cost of one pen = Rs 5.
*Graphical interpretation*: In both parts the straight lines representing the two equations intersect at a unique point; the coordinates of that point give the required numbers (boys‑girls) or costs (pencil‑pen).
1. Let \(b\) = number of boys and \(g\) = number of girls.
2. Total students: \(b+g = 10\) (Equation 1)
3. Girls are 4 more than boys: \(g = b + 4\) (Equation 2)
4. Graphical method – Plot the two lines on the \(b\)-\(g\) plane:
- For Eq. 1, take \(b=0\) ⇒ \(g=10\) and \(g=0\) ⇒ \(b=10\). Draw the line joining (0,10) and (10,0).
- For Eq. 2, take \(b=0\) ⇒ \(g=4\) and \(b=6\) ⇒ \(g=10\). Draw the line joining (0,4) and (6,10).
5. The two lines intersect at the point where both equations are satisfied. Solving algebraically gives:
\[g = b+4 \Rightarrow b+(b+4)=10 \Rightarrow 2b=6 \Rightarrow b=3\]
Substituting back: \(g = 3+4 = 7\).
6. Hence the intersection point is \((b,g) = (3,7)\).
Answer: 3 boys and 7 girls.
### Part (ii)
1. Let \(p\) = cost of one pencil (in rupees) and \(q\) = cost of one pen.
2. From the data:
- 5 pencils + 7 pens cost \(\text{Rs }50\): \(5p + 7q = 50\) (Equation 1)
- 7 pencils + 5 pens cost \(\text{Rs }46\): \(7p + 5q = 46\) (Equation 2)
3. Graphical method – Plot the two lines on the \(p\)-\(q\) plane:
- For Eq. 1, choose \(p=0\) ⇒ \(q=\frac{50}{7}\approx7.14\) and \(q=0\) ⇒ \(p=10\). Plot points (0,7.14) and (10,0).
- For Eq. 2, choose \(p=0\) ⇒ \(q=\frac{46}{5}=9.2\) and \(q=0\) ⇒ \(p=\frac{46}{7}\approx6.57\). Plot points (0,9.2) and (6.57,0).
4. The intersection of the two lines gives the simultaneous solution. Solving algebraically (as done for checking):
\[\begin{aligned}
5p+7q &= 50 \quad\text{(1)}\\
7p+5q &= 46 \quad\text{(2)}
\end{aligned}\]
Multiply (1) by 5 and (2) by 7:
\[25p+35q = 250\]
\[49p+35q = 322\]
Subtract: \(24p = 72 \Rightarrow p = 3\).
Substitute \(p=3\) into (1): \(5(3)+7q = 50 \Rightarrow 15+7q = 50 \Rightarrow 7q = 35 \Rightarrow q = 5\).
5. Hence the intersection point is \((p,q) = (3,5)\).
Answer: Cost of one pencil = Rs 3, cost of one pen = Rs 5.
*Graphical interpretation*: In both parts the straight lines representing the two equations intersect at a unique point; the coordinates of that point give the required numbers (boys‑girls) or costs (pencil‑pen).
Question 2
Hint available
On comparing the ratios 1 1 1 2 2 2 , and a b c a b c , find out whether the lines representing the following pairs of linear equations intersect at a point, are parallel or coincident: (i) 5x – 4y + 8 = 0 (ii) 9x + 3y + 12 = 0 7x + 6y – 9 = 0 18x + 6y + 24 = 0 (iii) 6x – 3y + 10 = 0 2x – y + 9 = 0
Key Idea
For two lines \(a_1x+b_1y+c_1=0\) and \(a_2x+b_2y+c_2=0\), compare the ratios \(\dfrac{a_1}{a_2},\;\dfrac{b_1}{b_2},\;\dfrac{c_1}{c_2}\):
- If \(\dfrac{a_1}{a_2}=\dfrac{b_1}{b_2}=\dfrac{c_1}{c_2}\) the lines are coincident.
- If \(\dfrac{a_1}{a_2}=\dfrac{b_1}{b_2}
eq\dfrac{c_1}{c_2}\) the lines are parallel (distinct).
- If the three ratios are all different, the lines intersect at a unique point.
- If \(\dfrac{a_1}{a_2}=\dfrac{b_1}{b_2}=\dfrac{c_1}{c_2}\) the lines are coincident.
- If \(\dfrac{a_1}{a_2}=\dfrac{b_1}{b_2}
eq\dfrac{c_1}{c_2}\) the lines are parallel (distinct).
- If the three ratios are all different, the lines intersect at a unique point.
Step-by-Step Solution
### Pair (i)
Only one equation \(5x-4y+8=0\) is given. To apply the ratio test we need two equations, therefore the nature of the pair cannot be decided with the information provided.
### Pair (ii)
We have two distinct equations:
\[
\begin{aligned}
&\text{(a)}\;9x+3y+12=0 \quad\Rightarrow\; a_1=9,\; b_1=3,\; c_1=12,\\[2mm]
&\text{(b)}\;7x+6y-9=0 \quad\Rightarrow\; a_2=7,\; b_2=6,\; c_2=-9.
\end{aligned}
\]
Step 1 – Form the ratios
\[
\frac{a_1}{a_2}=\frac{9}{7},\qquad \frac{b_1}{b_2}=\frac{3}{6}=\frac{1}{2},\qquad \frac{c_1}{c_2}=\frac{12}{-9}= -\frac{4}{3}.
\]
Step 2 – Compare
Since \(\frac{9}{7}
eq\frac{1}{2}
eq-\frac{4}{3}\), the three ratios are all different. Hence the two lines intersect at a single point.
Additional observation
The third equation \(18x+6y+24=0\) is exactly twice the first equation (multiply \(9x+3y+12=0\) by 2). Therefore the line represented by \(18x+6y+24=0\) is coincident with the line \(9x+3y+12=0\).
### Pair (iii)
Equations:
\[
\begin{aligned}
&\text{(a)}\;6x-3y+10=0 \quad\Rightarrow\; a_1=6,\; b_1=-3,\; c_1=10,\\[2mm]
&\text{(b)}\;2x-y+9=0 \quad\Rightarrow\; a_2=2,\; b_2=-1,\; c_2=9.
\end{aligned}
\]
Step 1 – Form the ratios
\[
\frac{a_1}{a_2}=\frac{6}{2}=3,\qquad \frac{b_1}{b_2}=\frac{-3}{-1}=3,\qquad \frac{c_1}{c_2}=\frac{10}{9}
eq3.
\]
Step 2 – Compare
Here \(\frac{a_1}{a_2}=\frac{b_1}{b_2}=3\) but \(\frac{c_1}{c_2}
eq3\). Hence the two lines are parallel (they have the same slope but different intercepts) and are not coincident.
Conclusion
- Pair (i): Insufficient data – cannot decide.
- Pair (ii): The lines \(9x+3y+12=0\) and \(7x+6y-9=0\) intersect at a point. The line \(18x+6y+24=0\) is coincident with \(9x+3y+12=0\).
- Pair (iii): The lines \(6x-3y+10=0\) and \(2x-y+9=0\) are parallel (distinct).
Only one equation \(5x-4y+8=0\) is given. To apply the ratio test we need two equations, therefore the nature of the pair cannot be decided with the information provided.
### Pair (ii)
We have two distinct equations:
\[
\begin{aligned}
&\text{(a)}\;9x+3y+12=0 \quad\Rightarrow\; a_1=9,\; b_1=3,\; c_1=12,\\[2mm]
&\text{(b)}\;7x+6y-9=0 \quad\Rightarrow\; a_2=7,\; b_2=6,\; c_2=-9.
\end{aligned}
\]
Step 1 – Form the ratios
\[
\frac{a_1}{a_2}=\frac{9}{7},\qquad \frac{b_1}{b_2}=\frac{3}{6}=\frac{1}{2},\qquad \frac{c_1}{c_2}=\frac{12}{-9}= -\frac{4}{3}.
\]
Step 2 – Compare
Since \(\frac{9}{7}
eq\frac{1}{2}
eq-\frac{4}{3}\), the three ratios are all different. Hence the two lines intersect at a single point.
Additional observation
The third equation \(18x+6y+24=0\) is exactly twice the first equation (multiply \(9x+3y+12=0\) by 2). Therefore the line represented by \(18x+6y+24=0\) is coincident with the line \(9x+3y+12=0\).
### Pair (iii)
Equations:
\[
\begin{aligned}
&\text{(a)}\;6x-3y+10=0 \quad\Rightarrow\; a_1=6,\; b_1=-3,\; c_1=10,\\[2mm]
&\text{(b)}\;2x-y+9=0 \quad\Rightarrow\; a_2=2,\; b_2=-1,\; c_2=9.
\end{aligned}
\]
Step 1 – Form the ratios
\[
\frac{a_1}{a_2}=\frac{6}{2}=3,\qquad \frac{b_1}{b_2}=\frac{-3}{-1}=3,\qquad \frac{c_1}{c_2}=\frac{10}{9}
eq3.
\]
Step 2 – Compare
Here \(\frac{a_1}{a_2}=\frac{b_1}{b_2}=3\) but \(\frac{c_1}{c_2}
eq3\). Hence the two lines are parallel (they have the same slope but different intercepts) and are not coincident.
Conclusion
- Pair (i): Insufficient data – cannot decide.
- Pair (ii): The lines \(9x+3y+12=0\) and \(7x+6y-9=0\) intersect at a point. The line \(18x+6y+24=0\) is coincident with \(9x+3y+12=0\).
- Pair (iii): The lines \(6x-3y+10=0\) and \(2x-y+9=0\) are parallel (distinct).
Question 3
Hint available
On comparing the ratios 1 1 2 2 , a b a b and 1 2 c c , find out whether the following pair of linear equations are consistent, or inconsistent. (i) 3x + 2y = 5 ; 2x – 3y = 7 (ii) 2x – 3y = 8 ; 4x – 6y = 9 (iii) 3 5 7 2 3 x y ; 9x – 10y = 14 (iv) 5x – 3y = 11 ; – 10x + 6y = –22 (v) 4 2 8 3 x y ; 2x + 3y = 12
Key Idea
For a pair of linear equations \(a_1x+b_1y=c_1\) and \(a_2x+b_2y=c_2\):
- If \(\dfrac{a_1}{a_2}=\dfrac{b_1}{b_2}=\dfrac{c_1}{c_2}\), the equations represent the same line – they are consistent (coincident).
- If \(\dfrac{a_1}{a_2}=\dfrac{b_1}{b_2}
eq\dfrac{c_1}{c_2}\), the lines are parallel and distinct – they are consistent (distinct).
- If \(\dfrac{a_1}{a_2}
eq\dfrac{b_1}{b_2}\), the lines intersect at a unique point – they are inconsistent (no common solution).
- If \(\dfrac{a_1}{a_2}=\dfrac{b_1}{b_2}=\dfrac{c_1}{c_2}\), the equations represent the same line – they are consistent (coincident).
- If \(\dfrac{a_1}{a_2}=\dfrac{b_1}{b_2}
eq\dfrac{c_1}{c_2}\), the lines are parallel and distinct – they are consistent (distinct).
- If \(\dfrac{a_1}{a_2}
eq\dfrac{b_1}{b_2}\), the lines intersect at a unique point – they are inconsistent (no common solution).
Step-by-Step Solution
### (i) \(3x+2y=5\) and \(2x-3y=7\)
- \(\dfrac{a_1}{a_2}=\dfrac{3}{2}=1.5\)
- \(\dfrac{b_1}{b_2}=\dfrac{2}{-3}= -\dfrac{2}{3}\)
- Since \(\dfrac{a_1}{a_2}
eq\dfrac{b_1}{b_2}\), the pair is inconsistent.
### (ii) \(2x-3y=8\) and \(4x-6y=9\)
- \(\dfrac{a_1}{a_2}=\dfrac{2}{4}=\dfrac{1}{2}\)
- \(\dfrac{b_1}{b_2}=\dfrac{-3}{-6}=\dfrac{1}{2}\)
- \(\dfrac{c_1}{c_2}=\dfrac{8}{9}
eq\dfrac{1}{2}\)
- Here \(\dfrac{a_1}{a_2}=\dfrac{b_1}{b_2}
eq\dfrac{c_1}{c_2}\); therefore the equations are consistent (distinct) (parallel lines).
### (iii) \(\dfrac{3}{5}x+\dfrac{7}{2}y=3\) and \(9x-10y=14\)
- \(\dfrac{a_1}{a_2}=\dfrac{\frac{3}{5}}{9}=\dfrac{3}{45}=\dfrac{1}{15}\)
- \(\dfrac{b_1}{b_2}=\dfrac{\frac{7}{2}}{-10}= -\dfrac{7}{20}\)
- Since \(\dfrac{a_1}{a_2}
eq\dfrac{b_1}{b_2}\), the pair is inconsistent.
### (iv) \(5x-3y=11\) and \(-10x+6y=-22\)
- \(\dfrac{a_1}{a_2}=\dfrac{5}{-10}= -\dfrac{1}{2}\)
- \(\dfrac{b_1}{b_2}=\dfrac{-3}{6}= -\dfrac{1}{2}\)
- \(\dfrac{c_1}{c_2}=\dfrac{11}{-22}= -\dfrac{1}{2}\)
- All three ratios are equal; hence the equations are consistent (coincident) (the same line).
### (v) \(\dfrac{4}{2}x+\dfrac{8}{3}y=\text{(constant)}\) and \(2x+3y=12\)
- Simplify the first coefficients: \(\dfrac{4}{2}=2\).
- \(\dfrac{a_1}{a_2}=\dfrac{2}{2}=1\)
- \(\dfrac{b_1}{b_2}=\dfrac{\frac{8}{3}}{3}=\dfrac{8}{9}
eq1\)
- Since \(\dfrac{a_1}{a_2}
eq\dfrac{b_1}{b_2}\), the pair is inconsistent.
Summary of results
- (i) Inconsistent
- (ii) Consistent (distinct)
- (iii) Inconsistent
- (iv) Consistent (coincident)
- (v) Inconsistent
- \(\dfrac{a_1}{a_2}=\dfrac{3}{2}=1.5\)
- \(\dfrac{b_1}{b_2}=\dfrac{2}{-3}= -\dfrac{2}{3}\)
- Since \(\dfrac{a_1}{a_2}
eq\dfrac{b_1}{b_2}\), the pair is inconsistent.
### (ii) \(2x-3y=8\) and \(4x-6y=9\)
- \(\dfrac{a_1}{a_2}=\dfrac{2}{4}=\dfrac{1}{2}\)
- \(\dfrac{b_1}{b_2}=\dfrac{-3}{-6}=\dfrac{1}{2}\)
- \(\dfrac{c_1}{c_2}=\dfrac{8}{9}
eq\dfrac{1}{2}\)
- Here \(\dfrac{a_1}{a_2}=\dfrac{b_1}{b_2}
eq\dfrac{c_1}{c_2}\); therefore the equations are consistent (distinct) (parallel lines).
### (iii) \(\dfrac{3}{5}x+\dfrac{7}{2}y=3\) and \(9x-10y=14\)
- \(\dfrac{a_1}{a_2}=\dfrac{\frac{3}{5}}{9}=\dfrac{3}{45}=\dfrac{1}{15}\)
- \(\dfrac{b_1}{b_2}=\dfrac{\frac{7}{2}}{-10}= -\dfrac{7}{20}\)
- Since \(\dfrac{a_1}{a_2}
eq\dfrac{b_1}{b_2}\), the pair is inconsistent.
### (iv) \(5x-3y=11\) and \(-10x+6y=-22\)
- \(\dfrac{a_1}{a_2}=\dfrac{5}{-10}= -\dfrac{1}{2}\)
- \(\dfrac{b_1}{b_2}=\dfrac{-3}{6}= -\dfrac{1}{2}\)
- \(\dfrac{c_1}{c_2}=\dfrac{11}{-22}= -\dfrac{1}{2}\)
- All three ratios are equal; hence the equations are consistent (coincident) (the same line).
### (v) \(\dfrac{4}{2}x+\dfrac{8}{3}y=\text{(constant)}\) and \(2x+3y=12\)
- Simplify the first coefficients: \(\dfrac{4}{2}=2\).
- \(\dfrac{a_1}{a_2}=\dfrac{2}{2}=1\)
- \(\dfrac{b_1}{b_2}=\dfrac{\frac{8}{3}}{3}=\dfrac{8}{9}
eq1\)
- Since \(\dfrac{a_1}{a_2}
eq\dfrac{b_1}{b_2}\), the pair is inconsistent.
Summary of results
- (i) Inconsistent
- (ii) Consistent (distinct)
- (iii) Inconsistent
- (iv) Consistent (coincident)
- (v) Inconsistent
Question 4
Hint available
Which of the following pairs of linear equations are consistent/inconsistent? If consistent, obtain the solution graphically: (i) x + y = 5, 2x + 2y = 10 (ii) x – y = 8, 3x – 3y = 16 (iii) 2x + y – 6 = 0, 4x – 2y – 4 = 0 (iv) 2x – 2y – 2 = 0, 4x – 4y – 5 = 0
Key Idea
For a pair of linear equations \(a_1x+b_1y=c_1\) and \(a_2x+b_2y=c_2\):
- If \(\frac{a_1}{a_2}=\frac{b_1}{b_2}=\frac{c_1}{c_2}\), the equations represent the same straight line (consistent, infinitely many solutions).
- If \(\frac{a_1}{a_2}=\frac{b_1}{b_2}
eq\frac{c_1}{c_2}\), the lines are parallel and distinct (inconsistent, no solution).
- If \(\frac{a_1}{a_2}
eq\frac{b_1}{b_2}\), the lines intersect at a unique point (consistent, one solution).
- If \(\frac{a_1}{a_2}=\frac{b_1}{b_2}=\frac{c_1}{c_2}\), the equations represent the same straight line (consistent, infinitely many solutions).
- If \(\frac{a_1}{a_2}=\frac{b_1}{b_2}
eq\frac{c_1}{c_2}\), the lines are parallel and distinct (inconsistent, no solution).
- If \(\frac{a_1}{a_2}
eq\frac{b_1}{b_2}\), the lines intersect at a unique point (consistent, one solution).
Step-by-Step Solution
### (i) \(x+y=5\) and \(2x+2y=10\)
1. Write in standard form: \(a_1=1,\;b_1=1,\;c_1=5\) and \(a_2=2,\;b_2=2,\;c_2=10\).
2. Compute ratios:
$$\frac{a_1}{a_2}=\frac{1}{2},\quad \frac{b_1}{b_2}=\frac{1}{2},\quad \frac{c_1}{c_2}=\frac{5}{10}=\frac{1}{2}.$$
3. Since all three ratios are equal, the two equations represent the same line. Hence the pair is consistent (dependent) – infinitely many solutions.
4. *Graphically*: both lines coincide; any point on the line \(x+y=5\) (e.g., \((0,5),(5,0)\)) satisfies both equations.
### (ii) \(x-y=8\) and \(3x-3y=16\)
1. Coefficients: \(a_1=1, b_1=-1, c_1=8\); \(a_2=3, b_2=-3, c_2=16\).
2. Ratios:
$$\frac{a_1}{a_2}=\frac{1}{3},\quad \frac{b_1}{b_2}=\frac{-1}{-3}=\frac{1}{3},\quad \frac{c_1}{c_2}=\frac{8}{16}=\frac{1}{2}.$$
3. Here \(\frac{a_1}{a_2}=\frac{b_1}{b_2}
eq\frac{c_1}{c_2}\). Hence the lines are parallel and distinct. The pair is inconsistent – no common solution.
4. *Graphically*: two parallel lines with slope 1, one passing through \((8,0)\) and the other through \((\frac{16}{3},0)\); they never meet.
### (iii) \(2x+y-6=0\) and \(4x-2y-4=0\)
1. Write as \(2x+y=6\) and \(4x-2y=4\). Coefficients: \(a_1=2, b_1=1, c_1=6\); \(a_2=4, b_2=-2, c_2=4\).
2. Ratios:
$$\frac{a_1}{a_2}=\frac{2}{4}=\frac{1}{2},\quad \frac{b_1}{b_2}=\frac{1}{-2}= -\frac{1}{2}.$$
Since \(\frac{a_1}{a_2}
eq\frac{b_1}{b_2}\), the lines intersect at a unique point – the pair is consistent (independent).
3. Solve algebraically:
\[\begin{aligned}
2x + y &= 6 \quad\text{(1)}\\
4x - 2y &= 4 \quad\text{(2)}
\end{aligned}\]
Multiply (1) by 2: \(4x + 2y = 12\).
Add to (2): \((4x-2y)+(4x+2y)=4+12\) ⇒ \(8x = 16\) ⇒ \(x = 2\).
Substitute in (1): \(2(2)+y=6\) ⇒ \(y=2\).
4. Solution: \((x,y) = (2,2)\).
5. *Graphically*: the line \(y = -2x + 6\) and the line \(y = 2x - 2\) intersect at the point \((2,2)\).
### (iv) \(2x-2y-2=0\) and \(4x-4y-5=0\)
1. Write as \(2x-2y=2\) and \(4x-4y=5\). Coefficients: \(a_1=2, b_1=-2, c_1=2\); \(a_2=4, b_2=-4, c_2=5\).
2. Ratios:
$$\frac{a_1}{a_2}=\frac{2}{4}=\frac{1}{2},\quad \frac{b_1}{b_2}=\frac{-2}{-4}=\frac{1}{2},\quad \frac{c_1}{c_2}=\frac{2}{5}.$$
3. Since \(\frac{a_1}{a_2}=\frac{b_1}{b_2}
eq\frac{c_1}{c_2}\), the lines are parallel and distinct. Hence the pair is inconsistent – no common solution.
4. *Graphically*: both lines have slope 1 (after rewriting as \(y = x -1\) and \(y = x -\frac{5}{4}\)), so they never intersect.
Summary of Results
- (i) Consistent – coincident lines (infinitely many solutions).
- (ii) Inconsistent – parallel distinct lines (no solution).
- (iii) Consistent – intersecting lines; solution \((2,2)\).
- (iv) Inconsistent – parallel distinct lines (no solution).
1. Write in standard form: \(a_1=1,\;b_1=1,\;c_1=5\) and \(a_2=2,\;b_2=2,\;c_2=10\).
2. Compute ratios:
$$\frac{a_1}{a_2}=\frac{1}{2},\quad \frac{b_1}{b_2}=\frac{1}{2},\quad \frac{c_1}{c_2}=\frac{5}{10}=\frac{1}{2}.$$
3. Since all three ratios are equal, the two equations represent the same line. Hence the pair is consistent (dependent) – infinitely many solutions.
4. *Graphically*: both lines coincide; any point on the line \(x+y=5\) (e.g., \((0,5),(5,0)\)) satisfies both equations.
### (ii) \(x-y=8\) and \(3x-3y=16\)
1. Coefficients: \(a_1=1, b_1=-1, c_1=8\); \(a_2=3, b_2=-3, c_2=16\).
2. Ratios:
$$\frac{a_1}{a_2}=\frac{1}{3},\quad \frac{b_1}{b_2}=\frac{-1}{-3}=\frac{1}{3},\quad \frac{c_1}{c_2}=\frac{8}{16}=\frac{1}{2}.$$
3. Here \(\frac{a_1}{a_2}=\frac{b_1}{b_2}
eq\frac{c_1}{c_2}\). Hence the lines are parallel and distinct. The pair is inconsistent – no common solution.
4. *Graphically*: two parallel lines with slope 1, one passing through \((8,0)\) and the other through \((\frac{16}{3},0)\); they never meet.
### (iii) \(2x+y-6=0\) and \(4x-2y-4=0\)
1. Write as \(2x+y=6\) and \(4x-2y=4\). Coefficients: \(a_1=2, b_1=1, c_1=6\); \(a_2=4, b_2=-2, c_2=4\).
2. Ratios:
$$\frac{a_1}{a_2}=\frac{2}{4}=\frac{1}{2},\quad \frac{b_1}{b_2}=\frac{1}{-2}= -\frac{1}{2}.$$
Since \(\frac{a_1}{a_2}
eq\frac{b_1}{b_2}\), the lines intersect at a unique point – the pair is consistent (independent).
3. Solve algebraically:
\[\begin{aligned}
2x + y &= 6 \quad\text{(1)}\\
4x - 2y &= 4 \quad\text{(2)}
\end{aligned}\]
Multiply (1) by 2: \(4x + 2y = 12\).
Add to (2): \((4x-2y)+(4x+2y)=4+12\) ⇒ \(8x = 16\) ⇒ \(x = 2\).
Substitute in (1): \(2(2)+y=6\) ⇒ \(y=2\).
4. Solution: \((x,y) = (2,2)\).
5. *Graphically*: the line \(y = -2x + 6\) and the line \(y = 2x - 2\) intersect at the point \((2,2)\).
### (iv) \(2x-2y-2=0\) and \(4x-4y-5=0\)
1. Write as \(2x-2y=2\) and \(4x-4y=5\). Coefficients: \(a_1=2, b_1=-2, c_1=2\); \(a_2=4, b_2=-4, c_2=5\).
2. Ratios:
$$\frac{a_1}{a_2}=\frac{2}{4}=\frac{1}{2},\quad \frac{b_1}{b_2}=\frac{-2}{-4}=\frac{1}{2},\quad \frac{c_1}{c_2}=\frac{2}{5}.$$
3. Since \(\frac{a_1}{a_2}=\frac{b_1}{b_2}
eq\frac{c_1}{c_2}\), the lines are parallel and distinct. Hence the pair is inconsistent – no common solution.
4. *Graphically*: both lines have slope 1 (after rewriting as \(y = x -1\) and \(y = x -\frac{5}{4}\)), so they never intersect.
Summary of Results
- (i) Consistent – coincident lines (infinitely many solutions).
- (ii) Inconsistent – parallel distinct lines (no solution).
- (iii) Consistent – intersecting lines; solution \((2,2)\).
- (iv) Inconsistent – parallel distinct lines (no solution).
Question 5
Hint available
Half the perimeter of a rectangular garden, whose length is 4 m more than its width, is 36 m. Find the dimensions of the garden.
Key Idea
Translate the word problem into a pair of linear equations in two variables (width and length) and solve them using substitution or elimination.
Step-by-Step Solution
1. Let the width of the garden be $w$ metres.
2. Since the length is 4 m more than the width, let the length be $l = w + 4$ metres.
3. Perimeter of a rectangle = $2(l + w)$. Half of the perimeter is therefore $\frac{1}{2}\times 2(l + w) = l + w$.
4. According to the statement, half the perimeter equals 36 m, so
$$l + w = 36.$$
5. Substitute $l = w + 4$ into the above equation:
$$(w + 4) + w = 36$$
$$2w + 4 = 36$$
6. Solve for $w$:
$$2w = 36 - 4 = 32$$
$$w = \frac{32}{2} = 16 \text{ m}$$
7. Find the length using $l = w + 4$:
$$l = 16 + 4 = 20 \text{ m}$$
8. Hence, the dimensions of the garden are:
- Width = 16 m
- Length = 20 m
2. Since the length is 4 m more than the width, let the length be $l = w + 4$ metres.
3. Perimeter of a rectangle = $2(l + w)$. Half of the perimeter is therefore $\frac{1}{2}\times 2(l + w) = l + w$.
4. According to the statement, half the perimeter equals 36 m, so
$$l + w = 36.$$
5. Substitute $l = w + 4$ into the above equation:
$$(w + 4) + w = 36$$
$$2w + 4 = 36$$
6. Solve for $w$:
$$2w = 36 - 4 = 32$$
$$w = \frac{32}{2} = 16 \text{ m}$$
7. Find the length using $l = w + 4$:
$$l = 16 + 4 = 20 \text{ m}$$
8. Hence, the dimensions of the garden are:
- Width = 16 m
- Length = 20 m
Question 6
Hint available
Given the linear equation 2x + 3y – 8 = 0, write another linear equation in two variables such that the geometrical representation of the pair so formed is: (i) intersecting lines (ii) parallel lines (iii) coincident lines
Key Idea
For a pair of linear equations \(a_1x+b_1y+c_1=0\) and \(a_2x+b_2y+c_2=0\):
- If \(\frac{a_1}{a_2}
eq \frac{b_1}{b_2}\) the lines intersect.
- If \(\frac{a_1}{a_2}=\frac{b_1}{b_2}
eq \frac{c_1}{c_2}\) the lines are parallel.
- If \(\frac{a_1}{a_2}=\frac{b_1}{b_2}=\frac{c_1}{c_2}\) the lines are coincident.
Thus by choosing appropriate coefficients for the second equation we can obtain each of the three required cases.
- If \(\frac{a_1}{a_2}
eq \frac{b_1}{b_2}\) the lines intersect.
- If \(\frac{a_1}{a_2}=\frac{b_1}{b_2}
eq \frac{c_1}{c_2}\) the lines are parallel.
- If \(\frac{a_1}{a_2}=\frac{b_1}{b_2}=\frac{c_1}{c_2}\) the lines are coincident.
Thus by choosing appropriate coefficients for the second equation we can obtain each of the three required cases.
Step-by-Step Solution
1. Given line : \(2x+3y-8=0\) \(\Rightarrow a_1=2,\; b_1=3,\; c_1=-8\).
2. Choose a second line \(a_2x+b_2y+c_2=0\) such that the ratios of the coefficients satisfy the required condition.
(i) Intersecting lines
- Choose coefficients so that \(\frac{a_1}{a_2}
eq\frac{b_1}{b_2}\).
- Example: take \(a_2=3,\; b_2=-2\). Then \(\frac{2}{3}
eq\frac{3}{-2}\).
- Choose any constant term, say \(c_2=5\).
- Hence the second equation is \(3x-2y+5=0\).
- Since the ratios of \(x\) and \(y\) coefficients are different, the two lines intersect at a unique point.
(ii) Parallel lines
- For parallelism we need \(\frac{a_1}{a_2}=\frac{b_1}{b_2}\) but \(\frac{c_1}{c_2}\) must be different.
- Multiply the given coefficients by a non‑zero constant, e.g. \(k=2\): \(a_2=2\times2=4,\; b_2=2\times3=6\).
- Keep the same ratio for \(c\) different from \(k\). Let \(c_2=-12\) (instead of \(-16\)).
- Second equation: \(4x+6y-12=0\).
- Here \(\frac{2}{4}=\frac{3}{6}=\frac{1}{2}\) but \(\frac{-8}{-12}=\frac{2}{3}
eq\frac{1}{2}\); therefore the lines are distinct and parallel.
(iii) Coincident lines
- For coincidence the three ratios must be equal.
- Use the same constant multiple for all coefficients, e.g. \(k=2\).
- Then \(a_2=4,\; b_2=6,\; c_2=-16\).
- Second equation: \(4x+6y-16=0\).
- Since \(\frac{2}{4}=\frac{3}{6}=\frac{-8}{-16}=\frac{1}{2}\), the two equations represent the same straight line; they are coincident.
3. Verification (optional):
- For (i) solving \(2x+3y-8=0\) and \(3x-2y+5=0\) simultaneously gives a unique solution, confirming intersection.
- For (ii) the slopes are \(-\frac{2}{3}\) for both lines, confirming parallelism.
- For (iii) the second equation is exactly twice the first, confirming coincidence.
2. Choose a second line \(a_2x+b_2y+c_2=0\) such that the ratios of the coefficients satisfy the required condition.
(i) Intersecting lines
- Choose coefficients so that \(\frac{a_1}{a_2}
eq\frac{b_1}{b_2}\).
- Example: take \(a_2=3,\; b_2=-2\). Then \(\frac{2}{3}
eq\frac{3}{-2}\).
- Choose any constant term, say \(c_2=5\).
- Hence the second equation is \(3x-2y+5=0\).
- Since the ratios of \(x\) and \(y\) coefficients are different, the two lines intersect at a unique point.
(ii) Parallel lines
- For parallelism we need \(\frac{a_1}{a_2}=\frac{b_1}{b_2}\) but \(\frac{c_1}{c_2}\) must be different.
- Multiply the given coefficients by a non‑zero constant, e.g. \(k=2\): \(a_2=2\times2=4,\; b_2=2\times3=6\).
- Keep the same ratio for \(c\) different from \(k\). Let \(c_2=-12\) (instead of \(-16\)).
- Second equation: \(4x+6y-12=0\).
- Here \(\frac{2}{4}=\frac{3}{6}=\frac{1}{2}\) but \(\frac{-8}{-12}=\frac{2}{3}
eq\frac{1}{2}\); therefore the lines are distinct and parallel.
(iii) Coincident lines
- For coincidence the three ratios must be equal.
- Use the same constant multiple for all coefficients, e.g. \(k=2\).
- Then \(a_2=4,\; b_2=6,\; c_2=-16\).
- Second equation: \(4x+6y-16=0\).
- Since \(\frac{2}{4}=\frac{3}{6}=\frac{-8}{-16}=\frac{1}{2}\), the two equations represent the same straight line; they are coincident.
3. Verification (optional):
- For (i) solving \(2x+3y-8=0\) and \(3x-2y+5=0\) simultaneously gives a unique solution, confirming intersection.
- For (ii) the slopes are \(-\frac{2}{3}\) for both lines, confirming parallelism.
- For (iii) the second equation is exactly twice the first, confirming coincidence.
Question 7
Hint available
Draw the graphs of the equations x – y + 1 = 0 and 3x + 2y – 12 = 0. Determine the coordinates of the vertices of the triangle formed by these lines and the x-axis, and shade the triangular region. 30
Key Idea
Find the intercepts of each line with the x‑axis, find the point of intersection of the two lines, and use these three points as vertices of the triangle.
Step-by-Step Solution
1. Write the equations in slope‑intercept form
$$\begin{aligned}
x-y+1&=0 \\Rightarrow y = x+1 \\[4pt]
3x+2y-12&=0 \\Rightarrow 2y = 12-3x \\Rightarrow y = 6-\frac{3}{2}x
\end{aligned}$$
2. Find the points where each line meets the x‑axis (y=0).
- For $y=0$ in $x-y+1=0$: $x+1=0 \Rightarrow x=-1$. Hence $A(-1,0)$.
- For $y=0$ in $3x+2y-12=0$: $3x-12=0 \Rightarrow x=4$. Hence $B(4,0)$.
3. Find the intersection of the two lines.
Set the right‑hand sides equal:
$$x+1 = 6-\frac{3}{2}x$$
$$\frac{5}{2}x =5 \Rightarrow x =2$$
Substituting in $y = x+1$ gives $y =3$. Hence $C(2,3)$.
4. Plot the three points $A(-1,0), B(4,0), C(2,3)$ on the coordinate plane and join them with straight lines. The lines $AB$, $AC$ and $BC$ are respectively the x‑axis, the line $x-y+1=0$ and the line $3x+2y-12=0$.
5. Shade the triangular region bounded by the two given lines and the x‑axis. This is the triangle $\triangle ABC$.
Thus the vertices of the required triangle are $(-1,0), (4,0), (2,3)$.
$$\begin{aligned}
x-y+1&=0 \\Rightarrow y = x+1 \\[4pt]
3x+2y-12&=0 \\Rightarrow 2y = 12-3x \\Rightarrow y = 6-\frac{3}{2}x
\end{aligned}$$
2. Find the points where each line meets the x‑axis (y=0).
- For $y=0$ in $x-y+1=0$: $x+1=0 \Rightarrow x=-1$. Hence $A(-1,0)$.
- For $y=0$ in $3x+2y-12=0$: $3x-12=0 \Rightarrow x=4$. Hence $B(4,0)$.
3. Find the intersection of the two lines.
Set the right‑hand sides equal:
$$x+1 = 6-\frac{3}{2}x$$
$$\frac{5}{2}x =5 \Rightarrow x =2$$
Substituting in $y = x+1$ gives $y =3$. Hence $C(2,3)$.
4. Plot the three points $A(-1,0), B(4,0), C(2,3)$ on the coordinate plane and join them with straight lines. The lines $AB$, $AC$ and $BC$ are respectively the x‑axis, the line $x-y+1=0$ and the line $3x+2y-12=0$.
5. Shade the triangular region bounded by the two given lines and the x‑axis. This is the triangle $\triangle ABC$.
Thus the vertices of the required triangle are $(-1,0), (4,0), (2,3)$.