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Maths 3 Mark Question

Linear Equations in Two Variables · Maharashtra Board STD 9 (MSBSHSE - English Medium). 18 questions.

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3 Mark Question

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18 questions · self-marked practice — reveal the answer and mark yourself.

Question 13 Marks
Find the function.

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$\therefore$ Given function $=\frac{6}{14}$
Verify the answer obtained.

Answer
For the fraction $\frac{6}{14}$, if the numerator is multiplied by 3 and 3 is subtracted from the denominator, we get fraction $\frac{18}{11}$.
Similarly, for the fraction $\frac{6}{14}$, if the numerator is increased by 8 and the denominator is doubled, we get fraction $\frac{1}{2}$.
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Question 23 Marks
There are instructions written near the arrows in the following diagram. From this information form suitable equations and write in the boxes indicated by arrows. Select any two equations from these boxes and find their solutions. Also verify the solutions. By taking one pair of equations at a time, how many pairs can be formed ? Discuss the solutions for these pairs.

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Answer
Here, if we take a pair of any two equations, we get following 6 pairs.

  1. equation (i) and (ii)
  2. equation (i) and (iii)
  3. equation (i) and (iv)
  4. equation (ii) and (iii)
  5. equation (ii) and (iv)
  6. equation (iii) and (iv)

Solution of each pair given above is (21, 15).
Here, all four equations are of same rectangle. By solving any two equations simultaneously, we get length and breadth of the rectangle.

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Question 33 Marks
On the glasses of following spectacles, write numbers such that
i. Their sum is 42 and difference is 16.

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ii. Their sum is 37 and difference is 11.

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iii. Their sum is 54 and difference is 20.

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iv. Their sum is … and difference is … .
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Answer
ii. x + y = 37 and x – y = 11
∴ x = 24, y = 13
iii. x +y = 54 and x – y = 20
∴ x = 37, y =17
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Question 43 Marks
By equating coefficients of variables, solve the following equations : 4x+y = 34 ; x + 4y = 16
Answer
4x + y = 34 …(i)
x + 4y = 16 …… (ii)
Multiplying equation (i) by 4,
16x + 4y = 136 …(iii)
Subtracting equation (ii) from (iii),

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x = 8
Substituting x = 8 in equation (i),
4x + y = 34
∴ 4(8) + y = 34
∴ 32 + y = 34
∴ y = 34 – 32 = 2
∴ (8, 2) is the solution of the given equations.

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Question 53 Marks
By equating coefficients of variables, solve the following equations : x – 2y = -10 ; 3x – 3y = -12
Answer
x – 2y = -10 ….(i)
3x – 5y = -12 …….(ii)
Multiplying equation (i) by 3,
3x – 6y = -30 …(iii)
Subtracting equation (ii) from (iii),

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∴ y = 18
Substituting y = 18 in equation (i),
x – 2y = -10
∴ x – 2(18) = -10
∴ x – 36 = -10
∴ x = -10 + 36 = 26
∴ (26, 18) is the solution of the given equations.

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Question 63 Marks
By equating coefficients of variables, solve the following equations : 5x + ly= 17 ; 3x – 2y = 4
Answer
5x + 7y = 17 …(i)
3x – 2y = 4 ….(ii)
Multiplying equation (i) by 2,
10x + 14y = 34 …(iii)
Multiplying equation (ii) by 7,
21x – 14y = 28 …..(iv)
Adding equations (iii) and (iv)

∴ x = 2
Substituting x = 2 in equation (ii),
3x – 2y = 4
∴ 3(2) – 2y = 4
∴ 6 – 2y = 4
∴ 6 – 4 = 2y
∴ 2 = 2y
$\therefore y=\frac{2}{2}$

∴ (2,1) is the solution of the given equations.

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Question 73 Marks
By equating coefficients of variables, solve the following equations $: 3x – 4y = 7 ; 5x + 2y = 3$
Answer
$3x – 4y = 7 …(i)$
$5x + 2y = 3 ….(ii)$
Multiplying equation (ii) by $2,$
$10x + 4y = 6 …(iii)$
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Adding equations $(i)$ and $(iii),$
$\therefore x = 1$
Substituting $x = 1$ in equation (i),
$3x – 4y = 7$
$\therefore 3(1) – 4y = 7$
$\therefore 3 – 4y = 7$
$\therefore 3 – 7 = 4y$
$\therefore -4 = 4y$
$ \therefore y=-\frac{4}{4}$
$\therefore y=-1 $
$\therefore(1,-1)$ is the solution of the given equations.
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Question 93 Marks
Solve the following simultaneous equations.
3x – 4y – 15 = 0 ........... (I)
y +x + 2 = 0 ........(II)
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Question 103 Marks
Sum of the ages of mother and son is 45 years. If son's age is subtracted from twice of mother's age then we get answer 54. Find the ages of mother and son. It becomes easy to solve a problem if we make use of variables.
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Question 113 Marks
3x – 4y – 15 = 0 and y + x + 2 = 0. Can these equations be solved by eliminating x ? Is the solution same?
Answer
3x – 4y – 15 = 0
∴ 3x – 4y = 15 …(i)
y + x + 2 = 0
∴ x + y = -2 ……(ii)
Multiplying equation (ii) by 3,
3x + 3y = -6 …(iii)
Subtracting equation (iii) from (i),

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∴ y = -3
Substituting y = -3 in equation (ii),
∴ x – 3 = -2
∴ x = – 2 + 3
∴ x = 1
∴ (x, y) = ( 1, -3)
Yes, the given equations can be solved by eliminating x. Also, the solution will remain the same.

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Question 123 Marks
x + y = 5 and 2x + 2y= 10 are two equations in two variables. Find live different solutions of x + y = 5, verify whether same solutions satisfy the equation 2x + 2y = 10 also. Observe both equations. Find the condition where two equations in two variables have all solutions in common.
Answer
Five solutions of x + y = 5 are given below:
(1,4), (2, 3), (3, 2), (4,1), (0, 5)
The above solutions also satisfy the equation 2x + 2y = 10.
∴ x + y = 5 …[Dividing both sides by 2]
∴ If the two equations are the same, then the two equations in two variables have all solutions common.
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Question 133 Marks
Solve the following sets of simultaneous equations : 2x – 7y = 7; 3x + y = 22
Answer
2x – 7y = 7 …(i)
3x + y = 22
∴ y = 22 – 3x ……(ii)
Substituting y = 22 – 3x in equation (i),
2x – 7(22 – 3x) = 7
∴ 2x – 154 + 21x = 7
∴ 23x = 7 + 154
∴ 23x = 161
$\therefore x=\frac{161}{23}$
∴ x = 7
Substituting x = 7 in equation (ii),
y = 22 – 3x
∴ y = 22 – 3(7)
∴ 7 = 22 -21= 1
∴ (7, 1) is the solution of the given equations.
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Question 143 Marks
Solve the following sets of simultaneous equations : 2x + 3y + 4 = 0; x – 5y = 11
Answer
2x + 3y + 4 = 0 …(i)
x – 5y = 11
∴x = 11 + 5y …(ii)
Substituting x = 11 + 5y in equation (i),
2(11 +5y) + 3y + 4 = 0
∴ 22 + 10y + 3y + 4 = 0
∴ 13y + 26 = 0
∴ 13y = -26
$\therefore y=\frac{-26}{13}$

∴ y = -2
Substituting y = -2 in equation (ii),
x = 11 + 5y
∴ x = 11 + 5(-2)
∴ x = 11 – 10 = 1
∴ (1, -2) is the solution of the given equations.

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Question 153 Marks
Solve the following sets of simultaneous equations : 2y – x = 0; 10x + 15y = 105
Answer
2y – x = 0
∴ x = 2y …(i)
10x + 15y = 105 …(ii)
Substituting x = 2y in equation (ii),
10(2y) + 15y = 105
∴ 20y + 15y = 105
∴ 35y = 105
$\therefore y =\frac{105}{35}$

∴ y = 3
Substituting y = 3 in equation (i),
x = 2y
∴ x = 2(3) = 6
∴ (6, 3) is the solution of the given equations.

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Question 163 Marks
Solve the following sets of simultaneous equations : 3x – 5y = 16; x – 3y= 8
Answer
3x – 5y = 16 …(i)
x – 3y = 8
∴x = 8 + 3y …..(ii)
Substituting x = 8 + 3y in equation (i),
3(8 + 3y) – 5y = 16
24 + 9y- 5y = 16
∴4y= 16 – 24
∴ 4y = -8
$\therefore y=\frac{-8}{4}$

y = -2
Substituting y = -2 in equation (ii),
x = 8 + 3 (-2)
∴ x = 8 – 6 = 2
∴ (2, -2) is the solution of the given equations.

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Question 173 Marks
Solve the following sets of simultaneous equations : 2x + y = 5 ; 3x – y = 5
Answer
2x + y = 5 …(i)
3x – y = 5 …(ii)
Adding equations (i) and (ii),
2x + y = 5
+ 3x – y = 5
5x = 10
$\therefore x=\frac{10}{5}$

∴ x = 2
Substituting x = 2 in equation (i),
2(2) + y = 5
4 + y = 5
∴ y = 5 – 4 = 1
∴ (2, 1) is the solution of the given equations.

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Question 183 Marks
Solve the following sets of simultaneous equations : x + y = 4 ; 2x – 5y = 1
Answer
Substitution Method:
x + y = 4
∴ x = 4 – y …(i)
2x – 5y = 1 ……(ii)
Substituting x = 4 – y in equation (ii),
2(4 – y) – 5y = 1
∴ 8 – 2y – 5y = 1
∴ 8 – 7y = 1
∴ 8 – 1 = 7y
∴ 7 = 7y
$\therefore y=\frac{7}{7}$

Substituting y = 1 in equation (i),
x = 4 – 1 = 3
∴ (3,1) is the solution of the given equations.

Alternate method:
Elimination Method:
x + y = 4 …(i)
2x – 5y = 1 ……(ii)
Multiplying equation (i) by 5,
5x + 5y = 20 … (iii)
Adding equations (ii) and (iii),
2x – 5y = 1
+ 5x + 5y = 20
7 = 21
$\therefore x=\frac{21}{7}$

∴ x = 3
Substituting x = 3 in equation (i),
3 + y = 4
∴ y = 4 – 3 = 1
(3,1) is the solution of the given equations.

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