Question 1
Solve the following pair of linear equations by the substitution method.
(i) x + y = 12 and x - y = 2
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Step 1: Express one variable
From x − y = 2, we get x = y + 2 ... (eq 1)
Step 2: Substitute
Substitute eq 1 into x + y = 12:
(y + 2) + y = 12
2y + 2 = 12 → 2y = 10 → y = 5
Step 3: Find x
x = 5 + 2 = 7
Answer: x = 7, y = 5
(ii) p - q = 2 and p/2 + q/4 = 3
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Step 1: Simplify
Multiply 2nd eq by 4: 2p + q = 12.
From 1st eq: p = q + 2.
Step 2: Substitute
2(q + 2) + q = 12
2q + 4 + q = 12 → 3q = 8 → q = 8/3
Step 3: Find p
p = 8/3 + 2 = 14/3
Answer: p = 14/3, q = 8/3
(iii) 2x - y = 2 and 4x - 2y = 4
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From eq 1: y = 2x − 2.
Substitute into eq 2:
4x − 2(2x − 2) = 4
4x − 4x + 4 = 4 → 4 = 4
Since the statement is true for all values, lines are coincident.
Answer: Infinitely many solutions.
(iv) 0.5x + 0.8y = 3.4 and 0.2x + 0.4y = 1.6
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Multiply both by 10:
5x + 8y = 34 ...(1)
2x + 4y = 16 → x + 2y = 8 → x = 8 − 2y ...(2)
Substitute (2) into (1):
5(8 − 2y) + 8y = 34
40 − 10y + 8y = 34 → −2y = −6 → y = 3
x = 8 − 2(3) = 2
Answer: x = 2, y = 3
(v) √3x + √2y = 0 and √2x - √8y = 0
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From eq 2: √2x = √8y → x = 2y.
Substitute into eq 1: √3(2y) + √2y = 0.
y(2√3 + √2) = 0 → y = 0.
Since x = 2y, then x = 0.
Answer: x = 0, y = 0
(vi) x/2 + 2y/3 = -1 and x - y/3 = 3
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From eq 2: x = 3 + y/3 = (9+y)/3.
Substitute into eq 1:
1/2 * (9+y)/3 + 2y/3 = −1
(9+y)/6 + 4y/6 = −1
9 + 5y = −6 → 5y = −15 → y = −3
x = 3 + (−3/3) = 2
Answer: x = 2, y = −3
Question 2
Solve 3x + 4y = 10 and 2x - 2y = 2, and find 'm' for which y = mx + 5.
Solution & Value of m
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Step 1: Solve for x and y
From 2x − 2y = 2 → x − y = 1 → x = y + 1.
Substitute into 3x + 4y = 10:
3(y + 1) + 4y = 10 → 7y = 7 → y = 1.
x = 1 + 1 = 2.
Step 2: Find m
Using y = mx + 5:
1 = m(2) + 5
2m = −4 → m = −2.
Answer: x=2, y=1, m=−2.
Question 3
Form the pair of linear equations for the following problems and find their solution by substitution method.
(i) The difference between two numbers is 20 and one number is five times the other. Find them.
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Let numbers be x and y.
x − y = 20 and x = 5y.
Substitute x: 5y − y = 20 → 4y = 20 → y = 5.
x = 5(5) = 25.
Answer: 25 and 5.
(ii) The larger of two supplementary angles exceeds the smaller by 20 degrees. Find them.
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x + y = 180 (Supplementary).
x = y + 20 (Difference).
Substitute: (y + 20) + y = 180 → 2y = 160 → y = 80°.
x = 80 + 20 = 100°.
Answer: 100° and 80°.
(iii) 3 bats and 4 balls cost ₹2500. 4 bats and 3 balls cost ₹2400. Find the cost of each.
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3x + 4y = 2500 ...(1)
4x + 3y = 2400 ...(2)
From (1): x = (2500 − 4y)/3. Sub into (2):
4(2500 − 4y)/3 + 3y = 2400
10000 − 16y + 9y = 7200
−7y = −2800 → y = 400.
x = (2500 − 1600)/3 = 300.
Answer: Bat: ₹300, Ball: ₹400.
(iv) Taxi charges: For 6 km cost is ₹58. For 10 km cost is ₹90. Find charges and cost for 20 km.
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Let fixed = x, per km = y.
x + 6y = 58 → x = 58 − 6y.
x + 10y = 90.
Substitute: (58 − 6y) + 10y = 90 → 4y = 32 → y = 8.
x = 58 − 48 = 10.
Cost for 20km: 10 + 20(8) = 170.
Answer: Fixed: ₹10, Per km: ₹8, Total: ₹170.
(v) A fraction becomes 4/5 if 1 is added to both numerator and denominator. It becomes 1/2 if 5 is subtracted from both. Find the fraction.
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(x+1)/(y+1) = 4/5 → 5x − 4y = −1.
(x−5)/(y−5) = 1/2 → 2x − y = 5 → y = 2x − 5.
Substitute y: 5x − 4(2x − 5) = −1
5x − 8x + 20 = −1 → −3x = −21 → x = 7.
y = 2(7) − 5 = 9.
Answer: 7/9.
(vi) 10 years hence, A will be twice as old as B. 10 years ago, A was 6 times as old as B. Find their present ages.
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A + 10 = 2(B + 10) → A = 2B + 10.
A − 10 = 6(B − 10) → A − 6B = −50.
Substitute A: (2B + 10) − 6B = −50.
−4B = −60 → B = 15.
A = 2(15) + 10 = 40.
Answer: A = 40, B = 15.