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
Step-by-Step Solution
Key Concept: For a pair of linear equations \(a_1x+b_1y=c_1\) and \(a_2x+b_2y=c_2\):<br>- 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).<br>- If \(\frac{a_1}{a_2}=\frac{b_1}{b_2}<br>eq\frac{c_1}{c_2}\), the lines are parallel and distinct (inconsistent, no solution).<br>- If \(\frac{a_1}{a_2}<br>eq\frac{b_1}{b_2}\), the lines intersect at a unique point (consistent, one 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).
Correct Answer: (i) Consistent (coincident); (ii) Inconsistent (parallel distinct); (iii) Consistent, intersect at (2,2); (iv) Inconsistent (parallel distinct).