Tag: graph of linear equations in two variables

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Multiple choice maths lines graphs of linear equations equations of lines parallel to the x-axis and y-axis graph of linear equations in two variables

If $2^{2x-y}=32$ and $2^{x+y}=16$ then $x^2+y^2$ is equal to

  1. 9

  2. 10

  3. 11

  4. 13
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

Given,

${ 2 }^{ 2x-y }=32$
${ 2 }^{ x+y }=16$
these equations can be writtn as
${ 2 }^{ \left( 2x-y \right)  }={ 2 }^{ 5 }$
$\Rightarrow 2x-y=5$.......................eq1
${ 2 }^{ x+y }={ 2 }^{ 4 }$
$\Rightarrow x+y=4$...........................eq2
on adding both equations
$ x+y=4$
$2x-y=5$
_______________________
$3x=9$
$x=3$
putting the value of x we get the value of y
$3+y=4$
$y=1$
putting the values of x and y we get
${ x }^{ 2 }+{ y }^{ 2 }$
${ 3 }^{ 2 }+{ 1 }^{ 2 }=9+1=10$

Multiple choice maths lines graphs of linear equations equations of lines parallel to the x-axis and y-axis graph of linear equations in two variables

If the expression $ \displaystyle (x+y)^{-1}. (x^{-1}+y^{-1})(xy^{-1}+x^{-1}y)^{-1} $ is simplefied it takes the form of which one of the following ?

  1. x+y
  2. $ \displaystyle( x^{2}+y^{2})^{-1}$

  3. xy
  4. $ \displaystyle x^{2}+y^{2}$
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation
$\left ( x+y \right )^{-1}\left ( \frac{1}{x}+\frac{1}{y} \right )^{-1}\left ( xy^{-1}+x^{-1}y \right )^{-1}$
$\frac{1}{x+y}\left ( \frac{1}{x}+\frac{1}{y} \right )\left ( \frac{x}{y}+\frac{y}{x}^{-1} \right )$
=$\frac{1}{x+y}\left ( \frac{x+y}{xy} \right )\left ( \frac{x^{2}+y^{2}}{xy} \right )^{-1}$
=$\frac{1}{x+y}\times \frac{x+y}{xy}\times \frac{xy}{x^{2}+y^{2}}$
=$\left ( \frac{1}{x^{2}+y^{2}} \right )=(x^{2}+y^{2})^{-1}$
Multiple choice maths lines graphs of linear equations equations of lines parallel to the x-axis and y-axis graph of linear equations in two variables

The angles between the lines $3 x + y - 7 = 0 \text { and } x + 2 y + 9 = 0$ is:

  1. $\dfrac { \pi } { 3 }$
  2. $\dfrac { \pi } { 6 }$
  3. $\dfrac { \pi } { 2 }$
  4. $\dfrac { \pi } { 4 }$
Reveal answer Fill a bubble to check yourself
D Correct answer
Explanation
Slope of the line $y=-3x+7$ is $-3$
Slope of the line $2y=-x-9$ or $y=\dfrac{-1}{2}x-\dfrac{9}{2}$ is $-\dfrac{1}{2}$

$\therefore\,{m} _{1}=-3,\,{m} _{2}=-\dfrac{1}{2}$

Angle between the line $=\tan{\theta}=\left|\dfrac{{m} _{1}-{m} _{2}}{1+{m} _{1}{m} _{2}}\right|$

$=\left|\dfrac{-3+\dfrac{1}{2}}{1+\left(-3\right)\left(-\dfrac{1}{2}\right)}\right|$

$=\left|\dfrac{\dfrac{-5}{2}}{1+\dfrac{3}{2}}\right|$

$=\left|\dfrac{\dfrac{-5}{2}}{\dfrac{2+3}{2}}\right|$

$=\left|\dfrac{\dfrac{-5}{2}}{\dfrac{5}{2}}\right|$

$\tan{\theta}=1$

$\therefore\,\theta=\dfrac{\pi}{4}$
Multiple choice maths lines graphs of linear equations equations of lines parallel to the x-axis and y-axis graph of linear equations in two variables

If the line segment joining $(2,3)$ and $(-1,2)$ is divided in the ratio $3:4$ by the graph of the equation $x+2y=k$, the value of $k$ is

  1. $\dfrac{5}{7}$
  2. $\dfrac{31}{7}$
  3. $\dfrac{36}{7}$
  4. $\dfrac{41}{7}$
Reveal answer Fill a bubble to check yourself
D Correct answer
Explanation

Using the section formula, find the coordinates of the point dividing the segment joining (2,3) and (-1,2) in the ratio 3:4. Substituting these coordinates into the equation x + 2y = k yields k = 41/7.

Multiple choice maths lines graphs of linear equations equations of lines parallel to the x-axis and y-axis graph of linear equations in two variables

If $4x+3y=120$, find how many non-negative integer solutions are possible?

  1. $1$
  2. $11$

  3. Infinite

  4. None of these
Reveal answer Fill a bubble to check yourself
D Correct answer
Explanation

We can write the equation in another form $4(x-3)+3(y+4)=120$
Now, we make a table

x 30 27 24 21 18 15 12 9 6 3 0
y 0 4 8 12 16 20 24 28 32 36 40

 We observe a patter that $x$ reduces by $3$ and $y$ increases by $4$. But both $x$ and $y$ cannot be negative or $0$. So, the value of $x=0$ and $y=0$ is ruled out. Hence, nine such positive integer solutions are possible.
Alternately, we can write the given equation as $4(x+3)+3(y-4)=120$. We get the same values of $x$ and $y$