Linear Equations Questions in English

(N/A) $(i)$ The equation $2x + 3y = 4.37$ can be written as $2x + 3y - 4.37 = 0$.
Comparing this with $ax + by + c = 0$,we get $a = 2$,$b = 3$,and $c = -4.37$.
$(ii)$ The equation $x - 4 = \sqrt{3}y$ can be written as $x - \sqrt{3}y - 4 = 0$.
Comparing this with $ax + by + c = 0$,we get $a = 1$,$b = -\sqrt{3}$,and $c = -4$.
$(iii)$ The equation $4 = 5x - 3y$ can be written as $5x - 3y - 4 = 0$.
Comparing this with $ax + by + c = 0$,we get $a = 5$,$b = -3$,and $c = -4$.
$(iv)$ The equation $2x = y$ can be written as $2x - y + 0 = 0$.
Comparing this with $ax + by + c = 0$,we get $a = 2$,$b = -1$,and $c = 0$.

(N/A) linear equation in two variables is expressed in the form $ax + by + c = 0$,where $a$,$b$,and $c$ are real numbers and $a$ and $b$ are not both zero.
$(i)$ $x = -5$ can be written as $1x + 0y = -5$,which is $1x + 0y + 5 = 0$.
$(ii)$ $y = 2$ can be written as $0x + 1y = 2$,which is $0x + 1y - 2 = 0$.
$(iii)$ $2x = 3$ can be written as $2x + 0y = 3$,which is $2x + 0y - 3 = 0$.
$(iv)$ $5y = 2$ can be written as $0x + 5y = 2$,which is $0x + 5y - 2 = 0$.

(A) Let the cost of a notebook be Rs. $x$.
Let the cost of a pen be Rs. $y$.
According to the given condition,the cost of a notebook is twice the cost of a pen.
Therefore,we can write the equation as:
$x = 2y$
Rearranging the terms to the standard form of a linear equation in two variables $(ax + by + c = 0)$:
$x - 2y = 0$
Thus,the required linear equation is $x - 2y = 0$.

(A) The given equation is $2x + 3y = 9.3\overline{5}$.
To express this in the standard form $ax + by + c = 0$,we subtract $9.3\overline{5}$ from both sides:
$2x + 3y - 9.3\overline{5} = 0$.
Comparing this with the general form $ax + by + c = 0$,we get:
$a = 2$
$b = 3$
$c = -9.3\overline{5}$

(A) The given equation is $-2x + 3y = 6$.
To express this in the form $ax + by + c = 0$,we subtract $6$ from both sides:
$-2x + 3y - 6 = 0$.
Comparing this with the standard form $ax + by + c = 0$,we identify the coefficients:
$a = -2$,
$b = 3$,
$c = -6$.

(N/A) The given equation is $2x = -5y$.
To express this in the form $ax + by + c = 0$,we move all terms to the left side:
$2x + 5y = 0$.
This can be written as $(2)x + (5)y + (0) = 0$.
Comparing this with the standard form $ax + by + c = 0$,we identify the coefficients as:
$a = 2$,$b = 5$,and $c = 0$.

(N/A) The given equation is $3x + 2 = 0$.
To express this in the form $ax + by + c = 0$,we can write it as:
$3x + 0y + 2 = 0$
Comparing this with the standard form $ax + by + c = 0$,we identify the coefficients as:
$a = 3$
$b = 0$
$c = 2$

(N/A) The given equation is $y - 2 = 0$.
To express this in the form $ax + by + c = 0$,we can rewrite it as:
$(0)x + (1)y + (-2) = 0$.
Comparing this with the standard form $ax + by + c = 0$,we identify the coefficients as:
$a = 0$,$b = 1$,and $c = -2$.

(A) The given equation is $5 = 2x$.
To express this in the form $ax + by + c = 0$,we rearrange the terms:
$2x - 5 = 0$
This can be written as:
$2x + 0y + (-5) = 0$
Comparing this with the standard form $ax + by + c = 0$,we identify the coefficients:
$a = 2$
$b = 0$
$c = -5$

(N/A) To draw the graph of the linear equation $x + y = 7$,we need at least two solutions of the equation.
$1$. If we put $x = 0$,then $0 + y = 7$,which gives $y = 7$. So,$(0, 7)$ is a solution.
$2$. If we put $x = 7$,then $7 + y = 7$,which gives $y = 0$. So,$(7, 0)$ is a solution.
We can represent these solutions in the following table:

$x$	$0$	$7$
$y$	$7$	$0$

Now,plot the points $(0, 7)$ and $(7, 0)$ on a Cartesian plane and join them with a straight line to obtain the graph of the equation $x + y = 7$.

(N/A) Here,the variables involved are force and acceleration. Let the force applied be $y$ units and the acceleration produced be $x$ units. From the concept of direct proportion,we can express this relationship as $y = kx$,where $k$ is a constant of proportionality. (From your study of science,you know that $k$ is actually the mass of the body.)
Since we do not know the specific value of $k$,we cannot draw a unique graph for $y = kx$. However,if we assign a specific value to $k$,we can draw the graph. Let us take $k = 3$,i.e.,we draw the line representing $y = 3x$.
To plot this,we find two solutions,for example,$(0, 0)$ and $(2, 6)$.
From the graph,you can see that when the force applied is $3$ units,the acceleration produced is $1$ unit. Also,note that $(0, 0)$ lies on the graph,which means that when the force applied is $0$ units,the acceleration produced is $0$ units.

(N/A) $x + y = 4$
$\Rightarrow y = 4 - x$
If we have $x = 0$,then $y = 4 - 0 = 4$.
If we have $x = 1$,then $y = 4 - 1 = 3$.
If we have $x = 2$,then $y = 4 - 2 = 2$.
$\therefore$ We get the following table:

$x$	$0$	$1$	$2$
$y$	$4$	$3$	$2$

Plot the ordered pairs $(0, 4)$,$(1, 3)$,and $(2, 2)$ on the graph paper. Joining these points,we get a line $AB$ as shown in the graph. Thus,the line $AB$ is the required graph of $x + y = 4$.

(N/A) $y=3x$
If $x=0$,then $y=3(0) \Rightarrow y=0$
If $x=1$,then $y=3(1) \Rightarrow y=3$
If $x=-1$,then $y=3(-1) \Rightarrow y=-3$
$\therefore$ We get the following table:

$x$	$0$	$1$	$-1$
$y$	$0$	$3$	$-3$

Plot the ordered pairs $(0, 0)$,$(1, 3)$,and $(-1, -3)$ on the graph paper. Joining these points,we get the straight line $LM$.
Thus,the line $LM$ is the required graph of $y=3x$.

(N/A) $y = 3 - 2x$
$\therefore$ If $x = 0$,then $y = 3 - 2(0) \Rightarrow y = 3$
If $x = 1$,then $y = 3 - 2(1) \Rightarrow y = 1$
If $x = 2$,then $y = 3 - 2(2) \Rightarrow y = -1$
$\therefore$ We get the following table:

$x$	$0$	$1$	$2$
$y$	$3$	$1$	$-1$

Plot the ordered pairs $(0, 3)$,$(1, 1)$,and $(2, -1)$ on the graph paper. Joining these points,we get a line $CD$.
Thus,the line $CD$ is the required graph of $3 = 2x + y$.

(N/A) Here,total distance covered $= x \, km$
Total taxi fare $= y$
Fare for the $1^{st} \, km = Rs. \, 8$
Remaining distance $= (x - 1) \, km$
$\therefore$ Fare for $(x - 1) \, km = Rs. \, 5(x - 1)$
Total taxi fare $= Rs. \, 8 + 5(x - 1)$
$\therefore$ According to the condition,
$y = 8 + 5(x - 1)$
$y = 8 + 5x - 5$
$y = 5x + 3$
This is the required linear equation representing the given information.
Graph: We have $y = 5x + 3$
When $x = 0, y = 5(0) + 3 = 3$
When $x = -1, y = 5(-1) + 3 = -2$
When $x = -2, y = 5(-2) + 3 = -7$
We get the following table:

$x$	$0$	$-1$	$-2$
$y$	$3$	$-2$	$-7$

Now,plotting the ordered pairs $(0, 3), (-1, -2)$ and $(-2, -7)$ on a graph paper and joining them,we get a straight line $PQ$. Thus,$PQ$ is the required graph of the linear equation $y = 5x + 3$.

(N/A) Constant force is $5$ units. Let the distance travelled $= x$ units and work done $= y$ units.
Since,$\text{Work done} = \text{Force} \times \text{Displacement}$
$\Rightarrow y = 5 \times x$
$\Rightarrow y = 5x$
Drawing the graph:
We have $y = 5x$.
When $x = 0$,then $y = 5(0) = 0$.
When $x = 1$,then $y = 5(1) = 5$.
When $x = 1.5$,then $y = 5(1.5) = 7.5$.
Therefore,we get the following table:

$x$	$0$	$1$	$1.5$
$y$	$0$	$5$	$7.5$

Plotting the ordered pairs $(0, 0)$,$(1, 5)$,and $(1.5, 7.5)$ on the graph paper and joining the points,we get a straight line.
From the graph:
$(i)$ When distance travelled $= 2$ units,then $x = 2$,which gives $y = 10$.
Therefore,$\text{Work done} = 10$ units.
$(ii)$ When distance travelled $= 0$ units,then $x = 0$,which gives $y = 5(0) = 0$.
Therefore,$\text{Work done} = 0$ units.

(N/A) Let the contribution of Yamini $= ₹ x$ and the contribution of Fatima $= ₹ y$.
$\therefore$ We have $x + y = 100$
$\Rightarrow y = 100 - x$
Now,we find the coordinates of points to plot the graph:
When $x = 0, y = 100 - 0 = 100$
When $x = 50, y = 100 - 50 = 50$
When $x = 100, y = 100 - 100 = 0$
We get the following table:

$x$	$0$	$50$	$100$
$y$	$100$	$50$	$0$

For drawing the graph,plot the ordered pairs $(0, 100)$,$(50, 50)$,and $(100, 0)$ on a graph paper. Joining these points,we get a line.
Thus,the line represents the graph of $x + y = 100$.

(N/A) $(i)$ We have the equation $F = \left(\frac{9}{5}\right) C + 32$.
When $C = 0$,$F = \left(\frac{9}{5}\right) \times 0 + 32 = 32$.
When $C = -15$,$F = \frac{9}{5}(-15) + 32 = -27 + 32 = 5$.
When $C = -10$,$F = \frac{9}{5}(-10) + 32 = 9(-2) + 32 = 14$.
Table of values:

$C$	$0$	$-15$	$-10$
$F$	$32$	$5$	$14$

Plotting the points $(0, 32)$,$(-15, 5)$,and $(-10, 14)$ on a graph and joining them gives a straight line.
$(ii)$ For $C = 30$,$F = \frac{9}{5}(30) + 32 = 9(6) + 32 = 54 + 32 = 86\,^oF$.
$(iii)$ For $F = 95$,$95 = \frac{9}{5}C + 32 \implies 63 = \frac{9}{5}C \implies C = 63 \times \frac{5}{9} = 35\,^oC$.
$(iv)$ For $C = 0$,$F = 32\,^oF$. For $F = 0$,$0 = \frac{9}{5}C + 32 \implies -32 = \frac{9}{5}C \implies C = -\frac{160}{9} \approx -17.8\,^oC$.
$(v)$ Let $F = C = x$. Then $x = \frac{9}{5}x + 32 \implies x - \frac{9}{5}x = 32 \implies -\frac{4}{5}x = 32 \implies x = 32 \times (-\frac{5}{4}) = -40$. Thus,$-40\,^oC = -40\,^oF$.

(N/A) $(i)$ $y = 3$ [An equation in one variable]
Since $y = 3$ is an equation in one variable,it represents a unique point on the number line.
The unique solution is a point at $3$ on the number line.
$(ii)$ $y = 3$ [An equation in two variables]
We can write $y = 3$ as $0x + y = 3$.
Now,for different values of $x$,we get $y = 3$:

$X$	$1$	$2$	$3$
$Y$	$3$	$3$	$3$

Plotting the ordered pairs $(1, 3)$,$(2, 3)$,and $(3, 3)$ on a Cartesian plane,we get a line $AB$ parallel to the $x$-axis,which represents the solution of $0x + y = 3$,i.e.,$y = 3$.

(A) linear equation in two variables is defined as an equation that can be written in the form $ax + by + c = 0$,where $a$,$b$,and $c$ are real numbers and $a$ and $b$ are not both zero.
This means that at least one of the coefficients of the variables $x$ or $y$ must be non-zero.
If $a = 0$ and $b = 0$,the equation becomes $c = 0$,which is not a linear equation in two variables.
Therefore,the condition for a linear equation in two variables is $a^2 + b^2 \neq 0$,which implies that $a$ and $b$ cannot both be zero simultaneously.

(A) In the Cartesian coordinate system,any point on the $x$-axis has a $y$-coordinate equal to $0$. Therefore,the equation representing all points on the $x$-axis is $y = 0$.

(FALSE) The statement is False.
By definition,a linear equation in two variables is of the form $ax + by + c = 0$,where $a, b,$ and $c$ are real numbers and $a$ and $b$ are not both zero. The geometric representation of such an equation in a Cartesian plane is always a straight line. Therefore,it is incorrect to say that the graph need not be a line.

(C) In a Cartesian coordinate system,any point is represented as $(x, y)$,where $x$ is the abscissa and $y$ is the ordinate.
According to the problem,the ordinate is $3$ times the abscissa.
Therefore,the relationship is $y = 3x$.
This is the required linear equation.

(N/A) $(i)$ $2x + 9 = 0$ [An equation in one variable]
We have: $2x + 9 = 0 \Rightarrow 2x = -9 \Rightarrow x = -4.5$ or $x = -\frac{9}{2}$.
This is a linear equation in one variable '$x$' only. Its solution is the point $-\frac{9}{2}$ on the number line.
$(ii)$ $2x + 9 = 0$ [An equation in two variables]
We can write $2x + 9 = 0$ as $2x + 0y + 9 = 0$ or $2x = -9 + 0y$.
Thus,$x = \frac{-9 + 0y}{2}$.
For any value of $y$,$x$ will always be $-\frac{9}{2}$.

$x$	$-\frac{9}{2}$	$-\frac{9}{2}$	$-\frac{9}{2}$
$y$	$1$	$2$	$3$

Plotting the points $\left(-\frac{9}{2}, 1\right)$,$\left(-\frac{9}{2}, 2\right)$,and $\left(-\frac{9}{2}, 3\right)$ on a Cartesian plane and joining them,we get a vertical line parallel to the $y$-axis.

(D) Let the total distance covered be $x$ $km$ and the total fare be ₹ $y$.
For the first $1$ $km$,the fare is ₹ $10$.
For the remaining distance $(x - 1)$ $km$,the fare is $3(x - 1)$.
Therefore,the total fare $y = 10 + 3(x - 1)$.
Simplifying the equation: $y = 10 + 3x - 3$,which gives $y = 3x + 7$.
To draw the graph,we find points:
If $x = 1, y = 10$.
If $x = 2, y = 13$.
If $x = 3, y = 16$.
Plotting these points on a graph and joining them gives a straight line.
For a journey of $4$ $km$,substitute $x = 4$ in the equation: $y = 3(4) + 7 = 12 + 7 = 19$.
Thus,the total fare for $4$ $km$ is ₹ $19$.

(N/A) The standard form of a linear equation in two variables is given by $ax + by + c = 0$,where $a$,$b$,and $c$ are real numbers and $a$ and $b$ are not both zero.
Given the equation: $5x - 3y = 15$.
To write it in standard form,subtract $15$ from both sides of the equation:
$5x - 3y - 15 = 0$.
Thus,the equation in standard form is $5x - 3y - 15 = 0$.

(B) The given equation is $3x = 10$.
Solving for $x$,we get $x = 10/3$.
This represents a vertical line passing through the point $(10/3, 0)$ on the $x$-axis.
Since the line is vertical,it is parallel to the $y$-axis.

(N/A) The standard form of a linear equation in two variables is $ax + by + c = 0$.
Given the equation: $3x + 4y = 24$.
To convert this to the standard form,subtract $24$ from both sides:
$3x + 4y - 24 = 0$.
Comparing this with $ax + by + c = 0$,we get:
$a = 3$
$b = 4$
$c = -24$.

(N/A) The given equation is $0.2x + 0.5y = 1.2$.
To write this in the standard form $ax + by + c = 0$,we subtract $1.2$ from both sides:
$0.2x + 0.5y - 1.2 = 0$.
Comparing this with the standard form $ax + by + c = 0$,we get:
$a = 0.2$
$b = 0.5$
$c = -1.2$

(N/A) The standard form of a linear equation in two variables is $ax + by + c = 0$.
Given the equation $2x = 3y$,we can rewrite it by bringing all terms to the left side:
$2x - 3y = 0$
This can be expressed as $2x - 3y + 0 = 0$.
Comparing this with the standard form $ax + by + c = 0$,we get:
$a = 2$
$b = -3$
$c = 0$

(N/A) linear equation in two variables is expressed in the form $ax + by + c = 0$.
Given the equation $2x = 9$,we can rewrite it as $2x + 0y = 9$.
Alternatively,it can be expressed as $2x + 0y - 9 = 0$.

(A) linear equation in two variables is expressed in the form $ax + by + c = 0$,where $a$,$b$,and $c$ are real numbers and $a$ and $b$ are not both zero.
Given the equation $4x - 17 = 0$,we can rewrite it by introducing the variable $y$ with a coefficient of $0$:
$4x + 0y - 17 = 0$.

(N/A) linear equation in two variables is expressed in the form $ax + by + c = 0$.
Given the equation $5 y = 3$,we can rewrite it by introducing the variable $x$ with a coefficient of $0$.
Thus,the equation becomes $0x + 5y = 3$.
Alternatively,it can be written in the standard form as $0x + 5y - 3 = 0$.

(N/A) linear equation in two variables is generally represented in the form $ax + by + c = 0$.
Given the equation $8y - 15 = 0$,we can express it in terms of two variables $x$ and $y$ by including the $x$ term with a coefficient of $0$.
Thus,the equation becomes $0x + 8y - 15 = 0$.

(N/A) To express the equation $5x + 3y = 45$ in the form $ax + by + c = 0$,we subtract $45$ from both sides:
$5x + 3y - 45 = 0$
Comparing this with the standard form $ax + by + c = 0$,we get:
$a = 5$
$b = 3$
$c = -45$

(N/A) To express the equation $0.3x = 0.8y - 2.4$ in the form $ax + by + c = 0$,we move all terms to the left side of the equation:
$0.3x - 0.8y + 2.4 = 0$
Comparing this with the standard form $ax + by + c = 0$,we get:
$a = 0.3$
$b = -0.8$
$c = 2.4$

(N/A) To express the given equation $x = \frac{2}{5}y + 10$ in the form $ax + by + c = 0$,we move all terms to the left side:
$x - \frac{2}{5}y - 10 = 0$
Comparing this with the standard form $ax + by + c = 0$,we get:
$a = 1$
$b = -\frac{2}{5}$
$c = -10$

(N/A) To express the equation in the form $ax + by + c = 0$,we subtract $18$ from both sides:
$\pi x - 3y - 18 = 0$
Comparing this with the standard form $ax + by + c = 0$,we get:
$a = \pi$
$b = -3$
$c = -18$

(N/A) To express the equation in the form $ax + by + c = 0$,we move all terms to the left side of the equation.
Given equation: $6x - 5y = 4.2\overline{7}$
Subtracting $4.2\overline{7}$ from both sides,we get: $6x - 5y - 4.2\overline{7} = 0$
Comparing this with the standard form $ax + by + c = 0$,we identify the coefficients:
$a = 6$
$b = -5$
$c = -4.2\overline{7}$

(N/A) To express the equation $2y - 3x = 14$ in the form $ax + by + c = 0$,we rearrange the terms:
$1$. Subtract $14$ from both sides: $2y - 3x - 14 = 0$.
$2$. Rearrange the terms to match the standard form $ax + by + c = 0$: $-3x + 2y - 14 = 0$.
$3$. Comparing this with $ax + by + c = 0$,we get:
$a = -3$
$b = 2$
$c = -14$

(N/A) To express the equation $3x = 2y + 1.\overline{5}$ in the form $ax + by + c = 0$,we move all terms to the left side:
$3x - 2y - 1.\overline{5} = 0$
Comparing this with the standard form $ax + by + c = 0$,we get:
$a = 3$
$b = -2$
$c = -1.\overline{5}$

(N/A) linear equation in two variables is represented in the form $ax + by + c = 0$.
Given the equation $5x = -10$,we can rewrite it as $5x + 10 = 0$.
To express this in two variables ($x$ and $y$),we include the $y$ term with a coefficient of $0$:
$5x + 0y + 10 = 0$.

(A) linear equation in two variables is expressed in the form $ax + by + c = 0$,where $a$,$b$,and $c$ are real numbers and $a$ and $b$ are not both zero.
Given the equation $\pi x = 3.14$,we can rewrite it by bringing all terms to one side:
$\pi x - 3.14 = 0$.
To express this in the form $ax + by + c = 0$,we include the $y$ term with a coefficient of $0$:
$\pi x + 0y - 3.14 = 0$.

(N/A) linear equation in two variables is represented in the form $ax + by + c = 0$.
Given the equation $4y = 24$,we can rewrite it as $0x + 4y - 24 = 0$.

(N/A) linear equation in two variables is represented in the form $ax + by + c = 0$.
Given the equation $3y = -11$,we can rewrite it as $0x + 3y + 11 = 0$.

(A) Let the price of $1$ sweetlime be $x$ and the price of $1$ orange be $y$ (in rupees).
According to the given statement,the total cost of $3$ sweetlimes and $6$ oranges is ₹ $210$.
Therefore,the linear equation in two variables is $3x + 6y = 210$.

(N/A) Let the price of a bat be $x$ and the price of a ball be $y$.
According to the problem,the price of a bat is five times the price of a ball.
Therefore,the linear equation in two variables is $x = 5y$ or $x - 5y = 0$.