Solution

The given linear equation $2x-5y=7$ has two variable. Since, a linear equation with two variable has infinitely many solution.

Hence, the correct option is (C).

Solution

The given equation $2x+5y=7$ has a unique solution, if $x,y$ are natural number.

Hence, the correct option is (a).

Solution

Put $x=2$ and $y=0$ in the linear equation $2x+3y=k$.

$2\times 2+3\times 0=k$

$k=4$

Hence, the correct option is (A).

Solution

Consider the linear equation:

$2x+0y+9=0$

Now,

$2x=-9$

$x=-{\displaystyle \frac{9}{2}}$

Since, the coefficient of $y$ is 0 in the given equation. So, the solution can be given as $(-{\displaystyle \frac{9}{2}},m)$.

Hence, the correct option is (A).

Solution

The graph of the linear equation $2x+3y=6$ cuts the $y$-axis at the point where $x$-coordinate is zero.

$\begin{array}{rl}& \text{ So, put }x=0\text{ in }2x+3y=6\text{, get: }\\ & \begin{array}{rl}2x+3y& =6\\ 2\times 0+3\times y& =6\\ 3y& =6\\ y& =2\end{array}\end{array}$

So, the point is $(0,2)$.

Hence, the correct option is (D).

Solution

The equation $x=7$ in two variable can be written as $1x+0y=7$.

Hence, the correct option is (B).

Solution

We know that any point on the $x$-axis has its ordinate 0 . Since, the point will be $(x,0)$.

Hence, the correct option is (C).

Solution

Any point on the line $y=x$ is of the form $(a,a)$ because the line $y=x$ will have $x$ and $y$ coordinate same.

Hence, the correct option is (a).

Solution

The equation of $x$-axis is of the form $y=0$.

Hence, the correct option is (B).

Solution

The graph of $y=6$ is a line parallel to $x$-axis because it does not contain $x$.

Hence, the correct option is (A).

Solution

$x=5,y=2$ is a solution of the linear equation $x+y=7$ because $5+2=7$.

Hence, the correct option is (C).

Solution

The point $(-2,2)$ and $(2,-2)$ have $x$ and $y$-coordinate of opposite signs.

Now, the sum of the $x$ and $y$-coordinate is: $x+y=-2+2=0$ i.e., $y=-x$ will have $x$ and $y$ coordinate opposite signs and also satisfies the point $(0,0)$.

Hence, the correct option is (B).

Solution

We know that I quadrant has consists of all the point $(x,y)$ positive. So, the positive solution of the equation $ax+by+c=0$ lie in I quadrant.

Hence, the correct option is (A).

Solution

The graph of the linear equation $2x+3y=6$ is a line which meets the $x$-axis at the point where $y=0$. Now, put $y=0$ in the given equation, get:

$2x+3y=6$

$2x+y\times 0=6$

$2x=6$

$x=3$

So, the point is $(3,0)$ which is lies on the line $2x+3y=6$.

Hence, the correct option is (C).

Solution

The graph of the linear equation $y=x$ passes through the point where $x$ and $y$ coordinate will have same that is $(1,1)$.

Hence, the correct option is (C).

Solution

If we multiply or divide both sides of a linear equation with a non-zero number, then the solution of the linear equation remains the same.

Solution

There are infinitely linear equations in $x$ and $y$ can be satisfied by $x=1$ and $y=2$. For example: $x+y=3$ or $2y-x=3$.

Hence, the correct option is $(C)$.

Solution

The point of the form $(a,a)$ always lies on $y=x$, because $x$ and $y$ coordinate are same. Hence, the correct option is (C).

Solution

The point of the form $(a,-a)$ always lies on the line $y=-x$ or $x+y=0$ because $x$ and $y$ coordinate have opposite signs.

Hence, the correct option is (D).

Solution

Consider the equation:

$3x+4y=12$

Putting $x=0$ and $y=3$ in the equation, get:

$3\times 0+4\times 3=12$

$12=12\text{ True }$

So, the point $(0,3)$ satisfies the equation $3x+4y=12$.

Hence, the given statement is true.

Solution

Consider the linear equation:

$x+2y=7$

Putting $x=0$ and $y=7$ in the given equation $x+2y=7$, get:

$x+2y=7$

$0+2\times 7=7$

$14=7$ Which is False

So, the point $(0,7)$ does not satisfy the equation.

Hence, the given statement is false.

Solution

The given equation is $x+y=0$ that is $y=-x$. Now, any point on the given graph has $x$ and $y$ -coordinate of opposite signs as, the point $(-1,1)$ and $(-3,3)$. So, these points satisfy the given equation.

Hence, the given statement is true.

Solution

See the given graph, the equation of the given line is $x=a$ that is $x=3$ parallel to the $y$-axis and to the right of $y$-axis, if $a>0$.

Hence, the given statement is true.

Solution

Consider the equation:

$x-y+2=0$.

The point $(0,2),(1,3),(2,4)$ and $(4,6)$ satisfy the given equation then they have solution of it. Now, the point $(3,-5)$ does not satisfy the given equation as $3-(-5)+2=0$, that is $3+5+2=0$ or $10=0$ that is false.

Hence, the given statement is false.

Solution

Every point on the graph of a linear equation in two variables does represent a solution of the linear equation. So, the given statement is false.

Solution

The graph of every linear equation in two variables is always a line. So, the given statement is false.

Solution

The point on the graph $y=x$ will have same signs of $x$ and $y$ coordinate but the point on the graph $y=-x$ will have opposite signs of $x$ and $y$ coordinate.

The graph of the linear equation $y=x$ and $y=-x$ on the same Cartesian plane is show below.

We observed that both graph are passes through the same point $(0,0)$ and they are mirror image of each other about the y-axis.

Solution

Consider the linear equation:

$2x+5y=19$

Let the coordinate of the point $(2,3)$.

Putting $x=2$ and $y=3$

$2x+5y=2\times 2+5\times 3$

$=4+15$

$=19$

Therefore, the point $(2,3)$ is the solution of the given equation.

Now, abscissa of the point is 2 and ordinate is 3 .

Then,

$2\times 1{\displaystyle \frac{1}{2}}=2\times {\displaystyle \frac{3}{2}}=3$

Hence, the ordinate of the point $(2,3)$ is $1{\displaystyle \frac{1}{2}}$ times its abscissa.

Solution

The graph of the equation $y=-3$ is parallel to the $x$-axis and at a distance 3 units and passing through the point $(0,-3)$ as show in the figure given below.

Solution

The graph of the linear equation whose solutions are represented by the points having the sum of the coordinates as 10 units is $x+y=10$.

When $x=0$ then $y=10$ and when $y=0$ then $x=10$.

Now, plot these two point $(0,10)$ and $(10,0)$ on the graph of the paper and joint them to get the straight line.

Hence, the graph of $x+y=10$ is a straight line as show in the figure given below.

Solution

The equation of linear equation such that point on its graph has an ordinate 3 times is $y=3x$.

Solution

Consider the linear equation:

$3y=ax+7$

The point $(3,4)$ lies on the graph of $3y=ax+7$. So, this point will be satisfy the given equation, get:

$3y=ax+7$

$12=3\times a+7$

$12-7=3a$

$3a=5$

$a={\displaystyle \frac{5}{3}}$

Hence, the value of $a$ is ${\displaystyle \frac{5}{3}}$.

Solution

(i) Consider the linear equation:

$2x+1=x-3$

$2x-x=-3-1$

$x=-4$

Hence, the $x=-4$ is the solution if the given equation in the number line.

(ii) The given linear equation $2x+1=x-3$ have infinitely many solution which are lies on the cartesion plane.

Solution

(i) The linear equation which lies on the $x$-axis has its ordinate 0 .

Now, Putting $y=0$ in the equation $x+2y=8$, get:

$\begin{array}{rl}x+2y& =8\\ x+2\times 0& =8\\ x& =8\end{array}$

(ii) The linear equation which lies on the y-axis has its abscissa 0 .

Now, Putting $x=0$ in the equation $x+2y=8$, get:

$x+2y=8$

$0+2y=8$

$\begin{array}{rl}& y={\displaystyle \frac{8}{2}}\\ & y=4\end{array}$

Solution

Consider the linear equation:

$2x+cy=8$

According to the question, when $x=y$. Putting $y=x$, get: $2x+cx=8$

$cx=8-2x$

$c={\displaystyle \frac{8-2x}{x}},x\ne 0$

Solution

Given:

$y$ varies directly as $x$

$y\propto x$

$y=kx$

Putting $y=12$, when $x=4$, get:

$12=k4$

$k={\displaystyle \frac{12}{4}}$

$k=3$

Therefore, the equation is $y=3x$.

Now, the value of $y$ when $x=5$ is $y=3\times 5=15$.

Solution

For $A(1,2)$, we have $2=9(1)-7=9-7=2$

For $B(-1,-16)$, we have $-16=9(-1)-7=-9-7=-16$

For $C(0,-7)$, we have $-7=9(0)-7=0-7=-7$

The line $y=9x-7$ is satisfied by the point A $(1,2),B(-1,-16)$ and C $(0,-7)$. Hence, the point $A(1,2),B(-1,-16)$ and $C(0,-7)$ are solution of the linear equation $y=9x-7$.

Solution

Consider the linear equation:

$3x+4y=6$

Both the point $(6,-2)$ and $(-6,6)$ satisfy the linear equation $3x+4y=6$. Now, plot the point $(6,-2)$ and $(-6,6)$ on the graph paper and then joint these two points and obtain a line. See the below graph it is cut the $x$-axis at point $(3,0)$ and $y$-axis at $(0,2)$.

Solution

Consider the linear equation:

$3x+4y=6$

The solution of given(above) linear equation cab be expressed as a table as follows:

$\begin{array}{|llll|}\hline x& 2& -2& 0\\ y& 0& 3& 1.5\\ \hline\end{array}$

Now, plot that above point on the graph paper and then join the points, see below:

Therefore, the graph cuts the $x$-axis at $(2,0)$ and $y$-axis at $(0,1.5)$.

Solution

Consider the linear equation:

$C={\displaystyle \frac{5F-160}{9}}$

(i) Putting $F={86}^{\circ }$, get:

$\begin{array}{rl}C& ={\displaystyle \frac{5F-160}{9}}\\ & ={\displaystyle \frac{5\times {86}^{\circ }-160}{9}}\\ & ={\displaystyle \frac{430-160}{9}}\\ & ={\displaystyle \frac{270}{9}}\\ & ={30}^{\circ }\end{array}$

Hence, the temperature in Celsius is ${30}^{\circ }$.

(ii) Putting $C={35}^{\circ }$, get:

$\begin{array}{rl}C& ={\displaystyle \frac{5F-160}{9}}\\ {35}^{\circ }& ={\displaystyle \frac{5F-160}{9}}\\ {315}^{\circ }& =5F-160\\ 5F& =315+160\\ & =415\\ F& ={\displaystyle \frac{475}{5}}\\ F& ={95}^{\circ }\end{array}$

Hence, the temperature in Fahrenheit is $95F$.

(iii) Putting $C=0^{\circ }$, get:

$\begin{array}{rl}C& ={\displaystyle \frac{5F-160}{9}}\\ 0& ={\displaystyle \frac{5F-160}{9}}\\ 0& =5F-160\\ 5F& =160\\ F& ={\displaystyle \frac{160}{5}}\\ F& ={32}^{\circ }\end{array}$

Now, putting $F=0^{\circ }$, get:

$\begin{array}{rl}C& ={\displaystyle \frac{5F-160}{9}}\\ C& ={\displaystyle \frac{5\times 0-160}{9}}\\ & =(-{\displaystyle \frac{160}{9}}{)}^{\circ }\end{array}$

If the temperature is $0F$, then the temperature in Fahrenheit is ${32}^{\circ }$ and if the temperature is $0F$ then the temperature in Celsius is $(-{\displaystyle \frac{160}{9}}{)}^{\circ }C$.

(iv) Putting $C=F$, get:

$\begin{array}{rl}C& ={\displaystyle \frac{5F-160}{9}}\\ C& =F\\ F& ={\displaystyle \frac{5F-160}{9}}\\ 9F-5F& =-160\\ 4F& =-160\\ F& ={\displaystyle \frac{-160}{4}}\\ F& =-{40}^{\circ }\end{array}$

Therefore, the numerical value of the temperature which is same in both the scales is -40 .

The linear equation that converts kelvin ( $x)$ to Fahrenheit (y) is given by the relation:

$y={\displaystyle \frac{9}{5}}(x-273)+32$

Solution

$y={\displaystyle \frac{9}{5}}(x-273)+32$

(i) When the temperature of the liquid is $x={313}^{\circ }K$.

$\begin{array}{rl}y& ={\displaystyle \frac{9}{5}}(313-273)+32\\ & ={\displaystyle \frac{9}{5}}\times 40+32\\ & ={72}^{\circ }+{32}^{\circ }\\ & ={104}^{\circ }F\end{array}$

(ii) When the temperature of the liquid is $y={158}^{\circ }F$.

$\begin{array}{rl}158& ={\displaystyle \frac{9}{5}}(x-273)+32\\ & ={\displaystyle \frac{9}{5}}(x-273)\\ & =158-32\\ x-273& =126\times {\displaystyle \frac{5}{9}}\\ & =70\\ x-273& =70\\ x& =273+70\\ x& ={343}^{\circ }K\end{array}$

Solution

According to the question:

$y\propto x$

$y=mx$

Now, make a table as follows by writing the values of $y$ below the corresponding value of $x$.

$\begin{array}{|llll|}\hline x& 0& 1& 2\\ y& 0& 6& 12\\ \hline\end{array}$

Plot the point $(0,0),(1,6)$ and $(2,12)$ on a graph paper. Now, joint all the points and obtain a line, see below.