🔶 EXERCISE 3.1 – NCERT SOLUTIONS
Q1. Is the point (2, 3) a solution of the equation 3x + y = 9?
Solution:
Substitute x=2x = 2x=2 and y=3y = 3y=3:3(2)+3=6+3=93(2) + 3 = 6 + 3 = 93(2)+3=6+3=9
✔ Yes, the point satisfies the equation.
✔ So (2, 3) is a solution.
Q2. Check whether (1, –1) satisfies 2x – 3y = 5.
Solution:
Put x=1,y=−1x = 1, y = -1x=1,y=−1:2(1)−3(−1)=2+3=52(1) – 3(-1) = 2 + 3 = 52(1)−3(−1)=2+3=5
✔ It satisfies the equation.
✔ So (1, –1) is a solution.
Q3. Determine whether points A(2,5), B(1,3), C(0,1) lie on the line 2x – y = –1.
Check each point:
For A(2,5):2(2)−5=4−5=−1✔2(2)-5 = 4-5 = -1 \quad ✔2(2)−5=4−5=−1✔
For B(1,3):2(1)−3=2−3=−1✔2(1)-3 = 2-3 = -1 \quad ✔2(1)−3=2−3=−1✔
For C(0,1):2(0)−1=−1✔2(0)-1 = -1 \quad ✔2(0)−1=−1✔
✔ All three points lie on the line.
🔶 EXERCISE 3.2 – NCERT SOLUTIONS
Q1. Solve by substitution:
x+y=14x + y = 14x+y=14 x−y=4x – y = 4x−y=4
Step 1: From 1st equation:x=14−yx = 14 – yx=14−y
Step 2: Substitute in 2nd equation:14−y−y=4⇒14−2y=4⇒2y=10⇒y=514 – y – y = 4 \Rightarrow 14 – 2y = 4 \Rightarrow 2y = 10 \Rightarrow y = 514−y−y=4⇒14−2y=4⇒2y=10⇒y=5
Step 3: Find x:x=14−5=9x = 14 – 5 = 9x=14−5=9
✔ Solution = (9, 5)
Q2. Solve:
3x+2y=13x + 2y = 13x+2y=1 2x−2y=−32x – 2y = -32x−2y=−3
Add both equations:(3x+2x)+(2y−2y)=1−3(3x + 2x) + (2y – 2y) = 1 – 3(3x+2x)+(2y−2y)=1−3 5x=−2⇒x=−255x = -2 \Rightarrow x = -\frac{2}{5}5x=−2⇒x=−52
Substitute in 2nd equation:2(−25)−2y=−32\left(-\frac{2}{5}\right) – 2y = -32(−52)−2y=−3 −45−2y=−3-\frac{4}{5} – 2y = -3−54−2y=−3 −2y=−3+45=−115-2y = -3 + \frac{4}{5} = -\frac{11}{5}−2y=−3+54=−511 y=1110y = \frac{11}{10}y=1011
✔ Solution = (−25,1110)\left(-\frac{2}{5}, \frac{11}{10}\right)(−52,1011)
🔶 EXERCISE 3.3 – NCERT SOLUTIONS (Elimination Method)
Q1. Solve the pair:
2x+3y=132x + 3y = 132x+3y=13 3x−2y=−23x – 2y = -23x−2y=−2
Multiply first by 3 and second by 2:6x+9y=396x + 9y = 396x+9y=39 6x−4y=−46x – 4y = -46x−4y=−4
Subtract:13y=43⇒y=4313=3.307…13y = 43 \Rightarrow y = \frac{43}{13} = 3.307…13y=43⇒y=1343=3.307…
Put y back:2x+3(4313)=132x + 3\left(\frac{43}{13}\right) = 132x+3(1343)=13
Solve:2x=13−12913=38−12913=−91132x = 13 – \frac{129}{13} = \frac{38 – 129}{13} = -\frac{91}{13}2x=13−13129=1338−129=−1391 x=−9126x = -\frac{91}{26}x=−2691
✔ Solution = (−9126,4313)\left(-\frac{91}{26}, \frac{43}{13}\right)(−2691,1343)
🔶 EXERCISE 3.4 – NCERT SOLUTIONS (Cross-Multiplication)
Q1. Solve:
6x−3y+10=06x – 3y + 10 = 06x−3y+10=0 2x−y+9=02x – y + 9 = 02x−y+9=0
Write in ax+by+c=0ax + by + c = 0ax+by+c=0 form:6x−3y+10=06x – 3y + 10 = 06x−3y+10=0 2x−y+9=02x – y + 9 = 02x−y+9=0
Apply cross-multiplication:x(−3)(9)−(−1)(10)=y(10)(2)−(9)(6)=16(−1)−(−3)(2)\frac{x}{(-3)(9) – (-1)(10)} = \frac{y}{(10)(2) – (9)(6)} = \frac{1}{6(-1) – (-3)(2)}(−3)(9)−(−1)(10)x=(10)(2)−(9)(6)y=6(−1)−(−3)(2)1
Solve each part:
Left:(−27)−(−10)=−17(-27) – (-10) = -17(−27)−(−10)=−17
Middle:20−54=−3420 – 54 = -3420−54=−34
Right:−6+6=0-6 + 6 = 0−6+6=0
Denominator is zero → No unique solution
⇒ Lines are parallel.
✔ No solution
🔶 EXERCISE 3.5 – NCERT SOLUTIONS (Word Problems)
Q1. Taxi charges:
Charge = ₹15 + ₹8 per km
Total cost for 10 km = ₹95
Let
xxx = fixed charge
yyy = rate per km
We get:
- x+10y=95x + 10y = 95x+10y=95
- x+y=23x + y = 23x+y=23
Subtract:9y=72⇒y=89y = 72 \Rightarrow y = 89y=72⇒y=8
Then:x=23−8=15x = 23 – 8 = 15x=23−8=15
✔ Fixed charge = ₹15
✔ Rate per km = ₹8