Question 12 Marks
Express the following equation in the form $ax + by + c = 0$ and indicate the values of $a, b, c$ in case.$\frac{\text{x}}{5}-\frac{\text{y}}{6}=1$
AnswerWe 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 $c = -30$
View full question & answer→Question 22 Marks
Express the following equation in the form $ax + by + c = 0$ and indicate the values of $a, b, c$ in case.$2\text{x}-\frac{\text{y}}{5}+6=0$
AnswerWe 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 $c = 30$
View full question & answer→Question 32 Marks
Express the following equation in the form $ax + by + c = 0$ and indicate the values of $a, b, c$ in case. $3y - 2x = 6$
AnswerWe have, $3y - 2x = 6 $
$\Rightarrow -2x + 3y - 6 = 0$
On comparing this equation with $ax + by + c = 0,$
we obtain $a = -2, b = 3$ and $c = -6$
View full question & answer→Question 42 Marks
Express the following equation in the form $ax + by + c = 0$ and indicate the values of $a, b, c$ in case. $3x - y = x - 1$
AnswerWe have, $3x - y = x - 1 $
$\Rightarrow 3x - x - y + 1 = 0 $
$\Rightarrow 2x - y + 1 = 0$
On comparing this equation with $ax + by + c = 0$, we obtain $a = 2, b = -1$ and $c = 1$
View full question & answer→Question 52 Marks
Check the following are the solution of the equation $5x - 4y = 20.\Big(2,\ \frac{-5}{2}\Big)$
AnswerThe 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.
View full question & answer→Question 62 Marks
Express the following equation in the form $ax + by + c = 0$ and indicate the values of $a, b, c$ in case. $3x + 5y = 7.5$
AnswerWe have, $3x + 5y = 7.5$
$\Rightarrow 3x + 5y - 7.5 = 0$
$\Rightarrow3\text{x}+5\text{y}-\frac{15}{2}=0$
$\Rightarrow 6x + 10y - 15 = 0$
On comparing this equation with $ax + by + c = 0$,
we obtain $a = 6, b = 10$ and $c = -15$
View full question & answer→Question 72 Marks
If $x = 3$ and $y = 4$ is a solution of the equation $5x - 3y = k,$ find the value of $k.$
AnswerGiven: $5x - 3y = k$ Since $x = 3$ and $y = 4$ is a solution of the given equation so, it should satisfy the equation.
$5(3) - 3(4) = k$
$\Rightarrow 15 - 12 = k$
$\Rightarrow 3 = k$
View full question & answer→Question 82 Marks
Find five different solution of the following equations:$\frac{2\text{x}}{5}+\frac{3\text{y}}{10}=3$
Answer$\frac{2\text{x}}{5}+\frac{3\text{y}}{10}=3$
|
$x$
|
$0$
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$\frac{15}{2}$
|
$5$
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$10$
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$3$
|
|
$y$
|
$10$
|
$0$
|
$\frac{10}{3}$
|
$\frac{-10}{3}$
|
$6$
|
View full question & answer→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 $k.$
AnswerGiven: $4x - 3y + 1 = 0 ...(1) x = 3k + 2$ and $y = 2k - 1$ Putting these values in the equation $(1),$
We get: $4(3k + 2) - 3(2k - 1) + 1 = 0 $
$\Rightarrow 12k + 8 - 6k + 3 + 1 = 0$
$ \Rightarrow 6k + 12 = 0 $
$\Rightarrow k + 2 = 0 $
$\Rightarrow k = -2$
View full question & answer→Question 102 Marks
Find five different solution of the following equations:$ 2x - 3y = 6$
Answer$2x - 3y = 6$
|
$x$
|
$0$
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$3$
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$-3$
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$\frac{9}{2}$
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$2$
|
|
$y$
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$-2$
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$0$
|
$-4$
|
$1$
|
$\frac{-2}{3}$
|
View full question & answer→Question 112 Marks
Express the following equation in the form $ax + by + c = 0$ and indicate the values of $a, b, c$ in case. $x = 6$
AnswerWe have, $x = 6$
$\Rightarrow x - 6 = 0 $
$\Rightarrow 1x + 0y - 6 = 0 $
$\Rightarrow x + 0y - 6 = 0$
On comparing this equation with $ax + by + c = 0$, we obtain $a = 1, b = 0 $and $c = -6$
View full question & answer→Question 122 Marks
Check the following are the solution of the equation $5x - 4y = 20. (0, -5)$
AnswerThe 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.
View full question & answer→Question 132 Marks
Express the following equation in the form $ax + by + c = 0$ and indicate the values of $a, b, c$ in case.$\sqrt{2}\text{x}+\sqrt{3}\text{y}=5$
AnswerWe 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 $c = -30$
View full question & answer→Question 142 Marks
Find five different solution of the following equations: $3y = 4x$
Answer$3y = 4x$
|
$x$
|
$3$
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$-3$
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$-6$
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$6$
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$0$
|
|
$y$
|
$4$
|
$-4$
|
$-8$
|
$8$
|
$0$
|
View full question & answer→Question 152 Marks
Express the following equation in the form $ax + by + c = 0$ and indicate the values of $a, b, c$ in case. $2x + 9 = 0$
AnswerWe have, $2x + 9 = 0 $
$\Rightarrow 2x + 0y + 9 = 0$ On comparing this equation with $ax + by + c = 0,$
we obtain $a = 2, b = 0$ and $c = 9$
View full question & answer→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. $y).$
AnswerLet: 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
$\Rightarrow 5x = 2y $
$\Rightarrow 5x - 2y = 0$
View full question & answer→Question 172 Marks
Check the following are the solution of the equation $5x - 4y = 20. (0, 5)$
AnswerThe 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.
View full question & answer→Question 182 Marks
Express the following equation in the form $a x+b y+c=0$ and indicate the values of $a, b, c$ in case. $4 x=5 y$
AnswerWe have, $4 x=5 y \Rightarrow 4 x-5 y=0$ On comparing this equation with $a x+b y+c=0$, we obtain $a=4, b=-5$ and $c=0$
View full question & answer→Question 192 Marks
Express the following equation in the form $ax + by + c = 0$ and indicate the values of $a, b, c$ in case.$\frac{\text{x}}{2}-\frac{\text{y}}{3}=\frac{1}{6}+\text{y}$
AnswerWe 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 $c = -1$
View full question & answer→Question 202 Marks
Express the following equation in the form $ax + by + c = 0$ and indicate the values of $a, b, c$ in case. $4y = 7$
AnswerWe have, $4y = 7$
$\Rightarrow 0x + 4y - 7 = 0$
On comparing this equation with $ax + by + c = 0$,
we obtain $a = 0, b = 4$ and $c = -7$
View full question & answer→Question 212 Marks
Express the following equation in the form $ax + by + c = 0$ and indicate the values of $a, b, c$ in case. $x + y = 4$
AnswerWe have, $x + y = 4 \Rightarrow x + y - 4 = 0$
On comparing this equation with $ax + by + c = 0$,
we obtain $a = 1, b = 1$ and $c = -4$
View full question & answer→Question 222 Marks
Check the following are the solution of the equation $5x - 4y = 20. (4, 0)$
AnswerThe 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.
View full question & answer→Question 232 Marks
Check the following are the solution of the equation $5x - 4y = 20$.$\Big(-2,\ \frac{5}{2}\Big)$
AnswerThe 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.
View full question & answer→