Mix Examples - Linear Equations in Two Variables Questions in English

(A) $(i)$ False,because $ax + by + c = 0$ is a linear equation in two variables only if both $a$ and $b$ are non-zero (i.e.,$a^2 + b^2 \neq 0$).
$(ii)$ False,because a linear equation in two variables has infinitely many solutions.
$(iii)$ False,the points $(2, 0)$ and $(-3, 0)$ lie on the $x$-axis because their $y$-coordinates are $0$. The point $(4, 2)$ lies in the first quadrant,and the point $(0, 5)$ lies on the $y$-axis.

(A) $(i)$ True. $A$ line parallel to the $y$-axis at a distance $a$ units to the left of the $y$-axis is represented by the equation $x = -a$. Since the distance is $4$ units,the equation is $x = -4$.
$(ii)$ False. For a graph to pass through the origin $(0, 0)$,the coordinates must satisfy the equation. Substituting $x = 0$ and $y = 0$ into $y = mx + c$ gives $0 = m(0) + c$,which implies $c = 0$. Since $c$ is not necessarily $0$ for all lines of the form $y = mx + c$,the statement is false.

(A) The given equation is $x + y = 0$,which can be rewritten as $y = -x$.
Any point on the graph of $y = -x$ must have $x$ and $y$ coordinates that are additive inverses of each other (i.e.,they have opposite signs and the same absolute value).
Looking at the graph,we can identify two points: $(-1, 1)$ and $(-3, 3)$.
For the point $(-1, 1)$,$x = -1$ and $y = 1$. Substituting these into the equation $x + y = 0$,we get $-1 + 1 = 0$,which is true.
For the point $(-3, 3)$,$x = -3$ and $y = 3$. Substituting these into the equation $x + y = 0$,we get $-3 + 3 = 0$,which is true.
Since both points satisfy the equation and two points determine a unique line,the given statement is True.

(N/A) Any point on the graph of $y = x$ will have $x$ and $y$ coordinates that are identical. The line passes through the points $(0,0), (1,1),$ and $(-1,-1).$
Similarly,any point on the graph of $y = -x$ will have $x$ and $y$ coordinates of opposite signs. The line passes through the points $(1,-1)$ and $(-1,1).$
Also,the point $(0,0)$ satisfies the equation $y = -x.$
The graphs of the linear equations $y = x$ and $y = -x$ on the same Cartesian plane are shown in the figure below.
We observe that the graphs of both these equations intersect at the origin $(0,0).$

(N/A) linear equation whose solutions are represented by the points having the sum of coordinates as $10$ units is $x+y=10$.
When $x=0$,$y=10$ and when $x=10$,$y=0$.
Now,plot these two points $(0, 10)$ and $(10, 0)$ on a graph paper and join them to obtain a straight line.
The graph of $x+y=10$ is a straight line as shown in the figure.

(N/A) $(i)$ The number of solution$(s)$ of the equation $2x + 1 = x - 3$ on the number line is $1$.
Solving the equation: $2x + 1 = x - 3 \Rightarrow 2x - x = -3 - 1 \Rightarrow x = -4$.
Thus,there is exactly one unique solution,$x = -4$,on the number line.
$(ii)$ The number of solution$(s)$ of the equation $2x + 1 = x - 3$ on the Cartesian plane is infinitely many.
In the Cartesian plane,the equation $2x + 1 = x - 3$ can be written as $x + 4 = 0$,which represents a vertical line parallel to the $y$-axis. Every point on this line $( -4, y )$ for any real value of $y$ satisfies the equation,hence there are infinitely many solutions.

(N/A) The given equation is $2x + 3y = 12$. To draw the graph of this equation,we need at least two points lying on the graph.
From the equation,we have $y = \frac{12 - 2x}{3}$.
For $x = 0$,$y = \frac{12 - 0}{3} = 4$. Therefore,$(0, 4)$ lies on the graph.
For $y = 0$,$2x = 12$,so $x = 6$. Therefore,$(6, 0)$ lies on the graph.
Now,plot the points $A(0, 4)$ and $B(6, 0)$ and join them to get the line $AB$.
Line $AB$ is the required graph. The graph (line $AB$) cuts the $x$-axis at the point $(6, 0)$ and the $y$-axis at the point $(0, 4)$.

(N/A) From the table,we get two points $A(1, 1)$ and $B(2, 3)$ which lie on the graph of the linear equation. Obviously,the graph will be a straight line. So,we first plot the points $A$ and $B$ and join them as shown in the figure.
To find the equation of the line passing through $(1, 1)$ and $(2, 3)$:
Slope $m = \frac{y_2 - y_1}{x_2 - x_1} = \frac{3 - 1}{2 - 1} = 2$.
Using the point-slope form $y - y_1 = m(x - x_1)$:
$y - 1 = 2(x - 1) \implies y - 1 = 2x - 2 \implies y = 2x - 1$.
$(i)$ To find where it cuts the $x$-axis,set $y = 0$:
$0 = 2x - 1 \implies 2x = 1 \implies x = \frac{1}{2}$.
So,it cuts the $x$-axis at the point $\left(\frac{1}{2}, 0\right)$.
$(ii)$ To find where it cuts the $y$-axis,set $x = 0$:
$y = 2(0) - 1 \implies y = -1$.
So,it cuts the $y$-axis at the point $(0, -1)$.

(N/A) Let the total distance covered be $x \text{ km}$ and the total fare charged be $Rs. y$.
For the first $1 \text{ km}$,the fare is $Rs. 10$.
For the remaining distance $(x - 1) \text{ km}$,the fare is $Rs. 4(x - 1)$.
Therefore,the total fare $y$ is given by:
$y = 10 + 4(x - 1)$
$y = 10 + 4x - 4$
$y = 4x + 6$
Thus,the required linear equation is $4x - y + 6 = 0$.
To draw the graph,we find two points on the line:
If $x = 0$,then $y = 4(0) + 6 = 6$. Point: $(0, 6)$.
If $x = -1$,then $y = 4(-1) + 6 = 2$. Point: $(-1, 2)$.
Plotting these points on a Cartesian plane and joining them gives the required straight line graph.

(N/A) Work done $= (\text{constant force}) \times (\text{distance})$
$= 3 \times (\text{distance})$
i.e., $y = 3x$, where $y$ (units) is the work done and $x$ (units) is the distance travelled.
Since $x = 2$ units (given), therefore, work done $= 3 \times 2 = 6$ units.
To plot the graph of the linear equation $y = 3x$, we need at least two solutions of the equation.
We see that $x = 0, y = 0$ satisfies the given equation, and $x = 1, y = 3$ also satisfies the equation.
Now we plot the points $A(0, 0)$ and $B(1, 3)$ and join them to form a line. The graph of the equation is a straight line. [We have not shown the whole line because work done cannot be negative].
To verify from the graph, draw a perpendicular to the $x$-axis at the point $(2, 0)$ meeting the graph at the point $C$. Clearly, the coordinates of $C$ are $(2, 6)$. It means that the work done is $6$ units.

(N/A) To find the linear equation $ax + by = c$,we use the two given points $(6, -2)$ and $(-6, 6)$.
For $(6, -2)$: $6a - 2b = c$
For $(-6, 6)$: $-6a + 6b = c$
Adding the two equations: $4b = 2c \implies c = 2b$.
Substituting $c = 2b$ into the first equation: $6a - 2b = 2b \implies 6a = 4b \implies a = \frac{2}{3}b$.
Let $b = 3$,then $a = 2$ and $c = 6$. The equation is $2x + 3y = 6$.
Plot the points $(6, -2)$ and $(-6, 6)$ on a Cartesian plane and join them with a straight line.
$(i)$ To find the $x$-intercept,set $y = 0$: $2x + 3(0) = 6 \implies 2x = 6 \implies x = 3$. The graph cuts the $x$-axis at $(3, 0)$.
$(ii)$ To find the $y$-intercept,set $x = 0$: $2(0) + 3y = 6 \implies 3y = 6 \implies y = 2$. The graph cuts the $y$-axis at $(0, 2)$.

(N/A) $C = \frac{5F - 160}{9}$
$(i)$ Putting $F = 86^{\circ}$,we get $C = \frac{5(86) - 160}{9} = \frac{430 - 160}{9} = \frac{270}{9} = 30^{\circ}$.
Hence,the temperature in Celsius is $30^{\circ} C$.
$(ii)$ Putting $C = 35^{\circ}$,we get $35 = \frac{5F - 160}{9} \Rightarrow 315 = 5F - 160$.
$\Rightarrow 5F = 315 + 160 = 475$.
$\therefore F = \frac{475}{5} = 95^{\circ}$.
Hence,the temperature in Fahrenheit is $95^{\circ} F$.
$(iii)$ Putting $C = 0^{\circ}$,we get $0 = \frac{5F - 160}{9} \Rightarrow 0 = 5F - 160$.
$\Rightarrow 5F = 160 \Rightarrow F = \frac{160}{5} = 32^{\circ}$.
Now,putting $F = 0^{\circ}$,we get $C = \frac{5(0) - 160}{9} = -\frac{160}{9} \approx -17.78^{\circ} C$.
If the temperature is $0^{\circ} C$,the temperature in Fahrenheit is $32^{\circ} F$,and if the temperature is $0^{\circ} F$,the temperature in Celsius is $-\frac{160}{9}^{\circ} C$.
$(iv)$ Putting $C = F$ in the given relation,we get $F = \frac{5F - 160}{9} \Rightarrow 9F = 5F - 160$.
$\Rightarrow 4F = -160 \Rightarrow F = -40^{\circ}$.
Hence,the numerical value of the temperature which is the same in both scales is $-40$.

(N/A) The given linear equation is $y = \frac{9}{5}(x - 273) + 32$.
$(i)$ Given $x = 313^{\circ} K$:
$y = \frac{9}{5}(313 - 273) + 32$
$y = \frac{9}{5}(40) + 32$
$y = 9 \times 8 + 32 = 72 + 32 = 104^{\circ} F$.
$(ii)$ Given $y = 158^{\circ} F$:
$158 = \frac{9}{5}(x - 273) + 32$
$158 - 32 = \frac{9}{5}(x - 273)$
$126 = \frac{9}{5}(x - 273)$
$x - 273 = 126 \times \frac{5}{9}$
$x - 273 = 14 \times 5 = 70$
$x = 273 + 70 = 343^{\circ} K$.

(N/A) Let $y$ be the force and $x$ be the acceleration. Since force is directly proportional to acceleration,we have $y \propto x$,which implies $y = mx$,where $m$ is the constant mass.
Given $m = 6 \,kg$,the linear equation is $y = 6x$.
To draw the graph,we find some points satisfying the equation:

$x$ (Acceleration)	$0$	$1$	$2$
$y$ (Force)	$0$	$6$	$12$

Plotting the points $(0,0)$,$(1,6)$,and $(2,12)$ on a graph and joining them gives a straight line.
From the graph:
$(i)$ When acceleration $x = 5 \,m/s^2$,the corresponding force $y = 30 \,N$.
$(ii)$ When acceleration $x = 6 \,m/s^2$,the corresponding force $y = 36 \,N$.

(D) To find the correct equation,we check which equation is satisfied by the points $(0,6)$,$(2,3)$,and $(4,0)$ shown on the graph.
$1$. For $x+y=4$: Substituting $(4,0)$ gives $4+0=4$ (True),but $(0,6)$ gives $0+6=6 \neq 4$ (False).
$2$. For $x+y=5$: Substituting $(2,3)$ gives $2+3=5$ (True),but $(4,0)$ gives $4+0=4 \neq 5$ (False).
$3$. For $x+y=6$: Substituting $(0,6)$ gives $0+6=6$ (True),but $(4,0)$ gives $4+0=4 \neq 6$ (False).
$4$. For $3x+2y=12$:
- For $(4,0)$: $3(4)+2(0) = 12+0 = 12$ (True).
- For $(2,3)$: $3(2)+2(3) = 6+6 = 12$ (True).
- For $(0,6)$: $3(0)+2(6) = 0+12 = 12$ (True).
Since all three points satisfy the equation $3x+2y=12$,this is the correct equation.

(A) We solve the equation $2y + 1 = y + 4$:
$\therefore 2y - y = 4 - 1$
$\therefore y = 3$
$(1)$ Representation on the number line:
Since $y = 3$ is an equation in one variable,it represents a unique point on the number line at the position $3$.
$(2)$ Representation on the Cartesian plane:
We know that $y = 3$ can be written as a linear equation in two variables: $0x + 1y = 3$.
For any value of $x$,the value of $y$ remains $3$. Thus,we can choose points like $(0, 3)$,$(1, 3)$,and $(3, 3)$. Plotting these points on the Cartesian plane and joining them results in a line parallel to the $x$-axis passing through $y = 3$.

(A) To find the intersection points with the coordinate axes:
- For the $x$-axis,set $y = 0$ and solve for $x$.
- For the $y$-axis,set $x = 0$ and solve for $y$.

Equation	Intersection at $x$-axis $(y=0)$	Intersection at $y$-axis $(x=0)$
$1. 3x + 5y = 15$	$(5, 0)$	$(0, 3)$
$2. 5x - 2y = 10$	$(2, 0)$	$(0, -5)$
$3. 4x + 3y = -12$	$(-3, 0)$	$(0, -4)$
$4. 3x - 7y = 21$	$(7, 0)$	$(0, -3)$
$5. x - y = 0$	$(0, 0)$	$(0, 0)$
$6. 2x - 3y = 0$	$(0, 0)$	$(0, 0)$
$7. x - y = -5$	$(-5, 0)$	$(0, 5)$
$8. 5x - 3y = 15$	$(3, 0)$	$(0, -5)$

(N/A) Step $1$: Solve the equation for $x$.
$3x - 1 = x + 7$
$3x - x = 7 + 1$
$2x = 8$
$x = 4$
Step $2$: Representation on the number line.
On the number line,mark the point $4$ as a solid dot.
Step $3$: Representation on the Cartesian plane.
In the Cartesian plane,the equation $x = 4$ represents a vertical line passing through $x = 4$ on the $x$-axis,parallel to the $y$-axis.

(Y = -3) Step $1$: Solve the equation for $y$.
$5y + 10 = 3y + 4$
$5y - 3y = 4 - 10$
$2y = -6$
$y = -3$
Step $2$: Representation on the number line.
On a horizontal number line,mark the point at $-3$.
Step $3$: Representation on the Cartesian plane.
The equation $y = -3$ represents a horizontal line parallel to the $x$-axis,passing through the point $(0, -3)$.

Let the coordinates of the point be $(x, y)$.
According to the problem,the sum of the coordinates is $10$,so the linear equation is $x + y = 10$.
To draw the graph,we find at least two solutions for the equation:
If $x = 0$,then $0 + y = 10 \implies y = 10$. So,the point is $(0, 10)$.
If $x = 10$,then $10 + y = 10 \implies y = 0$. So,the point is $(10, 0)$.
If $x = 5$,then $5 + y = 10 \implies y = 5$. So,the point is $(5, 5)$.
Plot these points $(0, 10)$,$(10, 0)$,and $(5, 5)$ on a Cartesian plane and join them with a straight line to obtain the graph of the equation $x + y = 10$.

(B) The equation is $x+y=11$.
To find the quadrants,we check the intercepts:
If $x=0$,then $y=11$. The point is $(0, 11)$,which lies on the $y$-axis between the first and second quadrants.
If $y=0$,then $x=11$. The point is $(11, 0)$,which lies on the $x$-axis between the first and fourth quadrants.
Since the line passes through $(0, 11)$ and $(11, 0)$,it passes through the first quadrant (where $x>0, y>0$),the second quadrant (where $x<0, y>0$ for $x < 0$),and the fourth quadrant (where $x>0, y<0$ for $x > 11$).
Thus,the line passes through the second,first,and fourth quadrants.