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Maths 2 Mark Question

Linear Equations in Two Variables · Maharashtra Board STD 9 (MSBSHSE - Semi English Medium). 23 questions.

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Question 12 Marks
Express the following equation in the form ax + by + c = 0 and indicate the values of a, b, c in case.$\frac{\text{x}}{5}-\frac{\text{y}}{6}=1$
Answer
We have,$\frac{\text{x}}{5}-\frac{\text{y}}{6}=1$
$\Rightarrow\frac{6\text{x}-5\text{y}}{30}=1$
$\Rightarrow6\text{x}-5\text{y}=30$
$\Rightarrow6\text{x}-5\text{y}-30=0$
On comparing this equation with ax + by + c = 0, we obtain a = 6, b = -5 and c = -30
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Question 22 Marks
Express the following equation in the form ax + by + c = 0 and indicate the values of a, b, c in case.$2\text{x}-\frac{\text{y}}{5}+6=0$
Answer
We have,$2\text{x}-\frac{\text{y}}{5}+6=0$
$\Rightarrow10\text{x}-\text{y}+30=0$
On comparing this equation with ax + by + c = 0, we obtain a = 10, b = -1 and c = 30
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Question 32 Marks
Express the following equation in the form ax + by + c = 0 and indicate the values of a, b, c in case.
3y - 2x = 6
Answer
We have,
3y - 2x = 6
⇒ -2x + 3y - 6 = 0
On comparing this equation with ax + by + c = 0, we obtain a = -2, b = 3 and c = -6
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Question 42 Marks
Express the following equation in the form ax + by + c = 0 and indicate the values of a, b, c in case.
3x - y = x - 1
Answer
We have,
3x - y = x - 1
⇒ 3x - x - y + 1 = 0
⇒ 2x - y + 1 = 0
On comparing this equation with ax + by + c = 0, we obtain a = 2, b = -1 and c = 1
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Question 52 Marks
Check the following are the solution of the equation 5x - 4y = 20.$\Big(2,\ \frac{-5}{2}\Big)$
Answer
The equation given is 5x - 4y = 20.$\Big(2,\ \frac{-5}{2}\Big)$
Putting the value in the given equation, We have: LHS $=5(2)-4\Big(\frac{-5}{2}\Big)$ = 10 + 10 = 20 RHS = 20 LHS = RHS Thus, $\Big(2,\ \frac{-5}{2}\Big)$ is a solution of the given equation.
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Question 62 Marks
Express the following equation in the form ax + by + c = 0 and indicate the values of a, b, c in case.
3x + 5y = 7.5
Answer
We have, 3x + 5y = 7.5 ⇒ 3x + 5y - 7.5 = 0$\Rightarrow3\text{x}+5\text{y}-\frac{15}{2}=0$
⇒ 6x + 10y - 15 = 0 On comparing this equation with ax + by + c = 0, we obtain a = 6, b = 10 and c = -15
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Question 72 Marks
If x = 3 and y = 4 is a solution of the equation 5x - 3y = k, find the value of k.
Answer
Given:
5x - 3y = k
Since x = 3 and y = 4 is a solution of the given equation so, it should satisfy the equation.
5(3) - 3(4) = k
⇒ 15 - 12 = k
⇒ 3 = k
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Question 82 Marks
Find five different solution of the following equations:$\frac{2\text{x}}{5}+\frac{3\text{y}}{10}=3$
Answer
$\frac{2\text{x}}{5}+\frac{3\text{y}}{10}=3$
x
0
$\frac{15}{2}$
5
10
3
y
10
0
$\frac{10}{3}$
$\frac{-10}{3}$
6
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Question 92 Marks
If x = 3k + 2 and y = 2k - 1 is a solution of the equation 4x - 3y + 1 = 0, find the value of k.
Answer
Given:
4x - 3y + 1 = 0 ...(1)
x = 3k + 2 and y = 2k - 1
Putting these values in the equation (1),
We get:
4(3k + 2) - 3(2k - 1) + 1 = 0
⇒ 12k + 8 - 6k + 3 + 1 = 0
⇒ 6k + 12 = 0
⇒ k + 2 = 0
⇒ k = -2
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Question 102 Marks
Find five different solution of the following equations:
2x - 3y = 6
Answer
2x - 3y = 6
x
0
3
-3
$\frac{9}{2}$
2
y
-2
0
-4
1
$\frac{-2}{3}$
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Question 112 Marks
Express the following equation in the form ax + by + c = 0 and indicate the values of a, b, c in case.
x = 6
Answer
We have,
x = 6
⇒ x - 6 = 0
⇒ 1x + 0y - 6 = 0
⇒ x + 0y - 6 = 0
On comparing this equation with ax + by + c = 0, we obtain a = 1, b = 0 and c = -6
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Question 122 Marks
Check the following are the solution of the equation 5x - 4y = 20.
(0, -5)
Answer
The equation given is 5x - 4y = 20.(0, -5)
Putting the value in the given equation,
We have:
LHS = 5(0) - 4(-5)
= 0 + 20
= 20
RHS = 20
LHS = RHS
Thus, (0, -5) is a solution of the given equation.
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Question 132 Marks
Express the following equation in the form ax + by + c = 0 and indicate the values of a, b, c in case.$\sqrt{2}\text{x}+\sqrt{3}\text{y}=5$
Answer
We have,$\sqrt{2}\text{x}+\sqrt{3}\text{y}=5$
$\sqrt{2}\text{x}+\sqrt{3}\text{y}-5=0$
On comparing this equation with ax + by + c = 0, we obtain $\text{a}=\sqrt{2},\ \text{b}=\sqrt{3}$ and c = -30
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Question 142 Marks
Find five different solution of the following equations:
3y = 4x
Answer
3y = 4x
x
3
-3
-6
6
0
y
4
-4
-8
8
0
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Question 152 Marks
Express the following equation in the form ax + by + c = 0 and indicate the values of a, b, c in case.
2x + 9 = 0
Answer
We have,
2x + 9 = 0
⇒ 2x + 0y + 9 = 0
On comparing this equation with ax + by + c = 0, we obtain a = 2, b = 0 and c = 9
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Question 162 Marks
The cost of 5 pencils is equal of the cost of 2 ballpoints. Write a linear equation in two variables to represent this statement. (Take the cost of a pencil to be Rs. x and that of a ballpoint to be Rs. y).
Answer
Let:
Cost of a pencil to be Rs. x and that of a ballpoint to be Rs y.
Cost of 5 pencils = 5x
Cost of 2 ballpoints = 2y
Cost of 5 pencils = Cost of 2 ballpoints
⇒ 5x = 2y
⇒ 5x - 2y = 0
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Question 172 Marks
Check the following are the solution of the equation 5x - 4y = 20.
(0, 5)
Answer
The equation given is 5x - 4y = 20. (0, 5) Putting the value in the given equation, We have: LHS = 5(0) - 4(5) = 0 - 20 = -20 RHS = 20$\text{LHS}\neq\text{RHS}$
Thus, (0, 5) is not a solution of the given equation.
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Question 182 Marks
Express the following equation in the form ax + by + c = 0 and indicate the values of a, b, c in case.
4x = 5y
Answer
We have,
4x = 5y
⇒ 4x - 5y = 0
On comparing this equation with ax + by + c = 0, we obtain a = 4, b = -5 and c = 0
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Question 192 Marks
Express the following equation in the form ax + by + c = 0 and indicate the values of a, b, c in case.$\frac{\text{x}}{2}-\frac{\text{y}}{3}=\frac{1}{6}+\text{y}$
Answer
We have,$\frac{\text{x}}{2}-\frac{\text{y}}{3}=\frac{1}{6}+\text{y}$
$\Rightarrow\frac{\text{x}}{2}-\frac{\text{y}}{3}-\text{y}=\frac{1}{6}$
$\Rightarrow\frac{3\text{x}-2\text{y}-6\text{y}}{6}=\frac{1}{6}$
$\Rightarrow3\text{x}-8\text{y}=1$
$\Rightarrow3\text{x}-8\text{y}-1=0$
On comparing this equation with ax + by + c = 0, we obtain a = 3, b = -8 and c = -1
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Question 202 Marks
Express the following equation in the form ax + by + c = 0 and indicate the values of a, b, c in case.
4y = 7
Answer
We have,
4y = 7
⇒ 0x + 4y - 7 = 0
On comparing this equation with ax + by + c = 0, we obtain a = 0, b = 4 and c = -7
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Question 212 Marks
Express the following equation in the form ax + by + c = 0 and indicate the values of a, b, c in case.
x + y = 4
Answer
We have,
x + y = 4
⇒ x + y - 4 = 0
On comparing this equation with ax + by + c = 0, we obtain a = 1, b = 1 and c = -4
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Question 222 Marks
Check the following are the solution of the equation 5x - 4y = 20.
(4, 0)
Answer
The equation given is 5x - 4y = 20.
(4, 0)
Putting the value in the given equation,
We have:
LHS = 5(4) - 4(0)
= 20
RHS = 20
LHS = RHS
Thus, (4, 0) is a solution of the given equation.
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Question 232 Marks
Check the following are the solution of the equation 5x - 4y = 20.$\Big(-2,\ \frac{5}{2}\Big)$
Answer
The equation given is 5x - 4y = 20.$\Big(-2,\ \frac{5}{2}\Big)$
Putting the value in the given equation, We have: LHS $=5(-2)-4\Big(\frac{5}{2}\Big)$ = -10 - 10 = -20 RHS = 20$\text{LHS}\neq\text{RHS}$
Thus, $\Big(-2,\ \frac{5}{2}\Big)$ is not a solution of the given equation.
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