1 Marks Question

Question 11 Mark

Solve the pair of linear equations by substitution method: $3x – y = 3; 9x – 3y = 9$

Answer

$3x - y = 3$
$9x - 3y = 9$
The given pair of linear equations is
$3x - y = 3..............(1)$
$9x - 3y = 9.............(2)$
From equation$(1),$
$y = 3x - 3...................(3)$
$9x - 3(3x - 3) = 9$
$\Rightarrow$ $9x - 9x + 9 = 9$
$\Rightarrow$ $9 = 9$
which is true. Therefore, equation $(1)$ and $(2)$ have infinitely many solutions.

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Question 21 Mark

Is the pair of linear equation consistent/inconsistent? If consistent, obtain the solution graphically: $2x – 2y – 2 = 0; 4x – 4y – 5 = 0$

Answer

$2 x - 2 x - 2 = 0................(1)$
$4 x - 4 y - 5 = 0..................(2)$
Here, $a _ { 1 } = 2 , \quad b = - 2 , c _ { 1 } = - 2$
$a _ { 2 } = 4 , b _ { 2 } = - 4 , c _ { 2 } = - 5$
We see that $\frac { a _ { 1 } } { a _ { 2 } } = \frac { b _ { 1 } } { b _ { 2 } } \neq \frac { c _ { 1 } } { c _ { 2 } }$
Hence, the lines represented by the equations$(1)$ and $( 2 )$ are parallel.
Therefore, equations $( 1)$ and $(2)$ have no solution, i.e., the given pair of a linear equation is inconsistent.

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Question 31 Mark

Is the pair of linear equation consistent/inconsistent? If consistent, obtain the solution graphically: $x – y = 8; 3x – 3y = 16$

Answer

$x - y = 8.................(1)$
$3 x - 3 y = 16.............(2)$
Here, $a _ { 1 } = 1 , b _ { 1 } = - 1 , c _ { 1 } = - 8$
$a _ { 2 } = 3 , b _ { 2 } = - 3 , c _ { 2 } = - 16$
We see that $\frac { a _ { 1 } } { a _ { 2 } } = \frac { b _ { 1 } } { b _ { 2 } } \neq \frac { c _ { 1 } } { c _ { 2 } }$
Hence, the lines represented by the equations$(1)$ and $(2)$ are parallel.
Therefore, equations $(1)$ and $( 2 )$ have no solution, i.e., the given pair of linear equation is inconsistent.

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Question 41 Mark

On comparing the ratios $ \frac { a _ { 1 } } { a _ { 2 } } , \frac { b _ { 1 } } { b _ { 2 } } $ and $\frac { c _ { 1 } } { c _ { 2 } }$, find out whether the pair of linear equation is consistent, or inconsistent: $5x - 3y = 11; -10x + 6y = -22$

Answer

From the given equations, We get,
$\frac{a_{1}}{a_{2}}=\frac{5}{-10}=-\frac{1}{2}$
$\frac{b_{1}}{b_{2}}=-\frac{3}{6}=-\frac{1}{2}$
$\frac{c_{1}}{c_{2}}=\frac{11}{-22}=-\frac{1}{2}$
Hence,
$\frac{a_{1}}{a_{2}}=\frac{b_{1}}{b_{2}}=\frac{c_{1}}{c_{2}}$
Therefore the given pair of line has an infinite number of solutions.
So the given pair of linear equation is consistent.

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Question 51 Mark

On comparing the ratios $\frac { a _ { 1 } } { a _ { 2 } } , \frac { b _ { 1 } } { b _ { 2 } } \text { and } \frac { c _ { 1 } } { c _ { 2 } }$, find out whether the pair of linear equations are consistent, or inconsistent: $\frac { 4 } { 3 } x$ $+ 2y = 8; 2x + 3y = 12.$

Answer

$$Given equations are:
$\frac { 4 } { 3 } x + 2y = 8; 2x + 3y = 12$
Compare equation $\frac{4}{3} x+2 y=8$ with $a_1 x+b_1 y+c_1=0$ and $2 x+3 y=12$
with $a_2 x+b_2 y+c_2=0$, We get, $a_1=\frac{4}{3}, a_1=\frac{4}{3}, b_1=2, c_1=-8, a_2=2, b_2=3, c_2=-12$
$\frac { a _ { 1 } } { a _ { 2 } } = \frac { \frac { 4 } { 3 } } { 2 } = \frac { 2 } { 3 } , \frac { b _ { 1 } } { b _ { 2 } } = \frac { 2 } { 3 }$ and $\frac { c _ { 1 } } { c _ { 2 } } = \frac { 8 } { 12 } = \frac { 2 } { 3 }$
Here $\frac { a _ { 1 } } { a _ { 2 } } = \frac { b _ { 1 } } { b _ { 2 } } = \frac { c _ { 1 } } { c _ { 2 } }$
Therefore, the lines have infinitely many solutions.
Hence, they are consistent.

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Question 61 Mark

Use elimination method to find all possible solutions of the following pair of linear equations:
$2x + 3y = 8 ...(1)$
$4x + 6y = 7 ...(2)$

Answer

Step $1$: Multiply equation $(i)$ by $2$ and equation $(ii)$ by $1$ to make the coefficients of $x$ equal. Then we get the equations as :
$4x + 6y = 16 ...(iii)$
$4x + 6y = 7 ...(iv)$
Step $2$: Subtracting equation $(iv)$ from equation $(iii)$,we get
$(4x - 4x) + (6y - 6y) = 16 - 7$
i.e., $0 = 9,$ which is a false statement. Therefore, the given pair of equations has no solution.

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Question 71 Mark

Two rails are represented by the equations: $x + 2y – 4 = 0$ and $2x + 4y – 12 = 0$. Will the rails cross each other?

Answer

The pair of linear equations are given as:
$x + 2y – 4 = 0 ...(i)$
$2x + 4y – 12 = 0 ...(ii)$
We express x in terms of $y$ from equation $(i)$, to get
$x = 4 – 2y$
Now, we substitute this value of $x$ in equation $(ii)$, to get
$2(4 – 2y) + 4y – 12 = 0$
$i.e., 8 – 12 = 0$
$i.e., – 4 = 0$
Which is a false statement. Therefore, the equations do not have a common solution.
So, the two rails will not cross each other.

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