Tag: reducing simple equations to simpler form

Questions Related to reducing simple equations to simpler form

Find the Solution  : $x - cy - bz = 0 $

                                 $ cx - y +az = 0 $
                                 $bx+ ay -z = 0 $

  1. $a^3 + b^3 + c^3 + 3abc = 1$

  2. $a^2 + b^2 + c^2 + 2abc = 1$

  3. $a^4 + b^4 + c^4 + 4abc = 1$

  4. $a^5 + b^5 + c^5 + 5abc = 1$


Correct Option: B
Explanation:

The given equations are 

$x-cy-bz = 0$
$cx - y + az = 0$
$bx + ay - z = 0$

Since x, y, z are not all zero, the system will have non-trivial solution if

$\triangle = \begin{vmatrix} 1 & -c & -b \ c & -1 & a \ b & a & -1\end{vmatrix} = 0$

or $1(1 - a^2) + c (-c -ab) -b (ac + b) = 0$

or $1 - a^2 - c^2 -abc - abc - b^2 = 0$

or $a^2 + b^2 + c^2 + 2abc = 1$

 Solve:$\dfrac{{7x - 2y}}{{xy}} = 5$ and $x=2y$

  1. $x=12/5 , y=6/5$

  2. $x=2/5 ,  y=1/5$

  3. $x=9 , y=9/2$

  4. $x=2 , y=1$


Correct Option: A
Explanation:
$\dfrac{7x-2y}{xy}=5$

$\Rightarrow 7x-2y=5xy$

$\Rightarrow \dfrac{7x}{xy}-\dfrac{2y}{xy}=5$

$\Rightarrow \dfrac{7}{y}-\dfrac{2}{x}=5$

Substitute $x=2y$ in the above equation, we get

$\dfrac{7}{y}-\dfrac{2}{2y}=5$

$\Rightarrow \dfrac{14-2}{2y}=5$

$\Rightarrow \dfrac{12}{2y}=5y$

$\Rightarrow 5y=6$

$\therefore y=\dfrac{6}{5}$

Put $y=\dfrac{6}{5}$ in $x=2y$ we get

$x=2\times \dfrac{6}{5}=\dfrac{12}{5}$

$\therefore x=\dfrac{12}{5}\,\,,y=\dfrac{6}{5}$

$\dfrac{2x-3}{2}-\dfrac{(x+1)}{3}=\dfrac{3x-8}{4}$

  1. $x=4$

  2. $x=-2$

  3. $x=2$

  4. $x=-4$


Correct Option: C
Explanation:

$\cfrac { 2x-3 }{ 2 } -\cfrac { x+1 }{ 3 } =\cfrac { 3x-8 }{ 4 } \ =>(6x-9-2x-2)\times 4=(3x-8)\times 6\ =>16x-44=18x-48\ =>2x=4\ =>x=2$

Solve: 

$3(x+1)=6$

  1. $x=-1$

  2. $x=\dfrac{1}{3}$

  3. $x=\dfrac{-1}{3}$

  4. $x={1}$


Correct Option: D
Explanation:
$3(x+1)=6$

$ \Rightarrow 3x+3=6$

$ \Rightarrow 3x=3$

$ \Rightarrow x=1$

The equation $x-\dfrac {8}{|x-3|}=3--\dfrac {8}{|x-3|}$ has

  1. Only one solution

  2. infinite solution

  3. no solution

  4. two solution


Correct Option: A
Explanation:

Consider the following equation.

$ x-\dfrac{8}{|x-3|}=3-\dfrac{8}{|x-3|} $

$ x-\dfrac{8}{\pm \left( x-3 \right)}=3-\dfrac{8}{\pm \left( x-3 \right)} $

$ x-\dfrac{8}{\left( x-3 \right)}=3-\dfrac{8}{\left( x-3 \right)} $

$ x=3 $

 $ x+\dfrac{8}{\left( x-3 \right)}=3+\dfrac{8}{\left( x-3 \right)} $

$ x=3 $

Hence, this is the correct answer.

If $t = x+2$, find the value of x .If $2t-7 +\dfrac{3(t-1)}{2}=3$

  1. $\frac{23}{7}$

  2. $\frac{3}{7}$

  3. $\frac{9}{7}$

  4. $\frac{37}{7}$


Correct Option: C
Explanation:

$(2t-7)+(\frac{3t-3}{2})=3\\(\frac{4t-14+3t-3}{2})=3\\7t-17=6\\\therefore 7t=6+17\\t=(\frac{23}{7})\\then x+2=(\frac{23}{7})\\\therefore x=(\frac{23}{7})-2\\=(\frac{23-14}{7})\\=(\frac{9}{7})$

Which equation is  non- linear 

  1. $x + 1 = 2x - 3$

  2. $9 - 5{x^2} = 4$

  3. $- 2x - 5 - \left( {x + 4} \right) = - 9$

  4. ${x} - \frac{1}{{15}} = 2$


Correct Option: B
Explanation:

Linear equations are those which have the degree of equation as $1$


Here every equation has the degree as $1$

Except $9-5x^2=4$

So its is not a linear equation 

A triangular number which is the sum of the square of two consecutive odd numbers is?

  1. $10$

  2. $15$

  3. $21$

  4. $28$


Correct Option: A
Explanation:
Let the two consecutive odd numbers be $n-2,n$
$(n-2)^2+n\\ n^2+4-4n+n^2\\2n^2+4-4n$
If we put $n=1\\ 2(1)-4+4=2\\n=2\\ 2(4)+4-4=4\\n=3\\2(9)+4-4(3)\\=18+4-12\\=10$
$\therefore $ Triangular no. is $10$
For any other value of $n$ the triangular no. is not in the required options
$\therefore$ we take $n=3$ and triangular no. is $10$

If $\displaystyle \frac{a}{3y}+\frac{3b}{x}=7$ and $\displaystyle a+1=2b+1=x=5,$ find the value of $'y'.$


  1. $\displaystyle \frac{10}{77}$

  2. $\displaystyle \frac{22}{69}$

  3. $\displaystyle \frac{20}{87}$

  4. $\displaystyle \frac{14}{93}$


Correct Option: C
Explanation:

Since, $ a+1=2b+1=x=5 $
$ => a = 5 - 1 = 4 $
$ 2b = 5 - 1 = 4 => b = 2 $

$ x = 5 $

Substituting these values in $ \frac { a }{ 3y } +\frac { 3b }{ x } =7 $, we get $ \frac { 4 }{ 3y } +\frac { 6 }{ 5 } =7 $
$ => \frac { 4 }{ 3y } = 7 - \frac { 6 }{ 5 } $
$ => \frac { 4 }{ 3y } =  \frac { 29 }{ 5 } $
$ => \frac { 3y }{ 4 } =  \frac { 5 }{ 29 } $
$ => y = \frac {20}{87} $

If $(2ax + 1) (3x + 1) = 6a (x + 1)$ and $x = 1$, find the value of $a$.

  1. $1$

  2. $4$

  3. $3$

  4. $2$


Correct Option: A
Explanation:

    $(2ax + 1) (3x + 1) = 6a (x + 1) $
$=>[2a(1)+1][3(1)+1]=6a[(1+1)]$
$=>(2a+1)(3+1)=6a(2)$
$=>(2a+1)(4)=12a$
$=>8a+4=12a$
$=>8a-12a=-4$
$=>-4a=-4$
$=>a=1$