Mathematics · Quantitative Aptitude

Linear Equations

196 Questions

Linear equations involve solving for unknown variables in single or multi variable systems. These questions test algebraic manipulation and logical consistency skills. They are a core component of quantitative aptitude and advanced mathematics tests.

Solving simultaneous equationsSingle variable equationsSystem consistency checksIndeterminate equationsMatrix form solutions

Linear Equations Questions

Multiple choice lines in planes applications of determinants inverse of a matrix and linear equations matrix algebra maths

To solve  $x + y = 3 : 3 x - 2 y - 4 = 0$  by determinant method find  $D.$

  1. $5$
  2. $1$
  3. $-5$
  4. $-1$
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation
${ a } _{ 1 }x+{ b } _{ 1 }y-{ c } _{ 1 }=0\quad \Rightarrow { a } _{ 1 }x+{ b } _{ 1 }y={ c } _{ 1 }$
${ a } _{ 2 }x+{ b } _{ 2 }y-{ c } _{ 2 }=0\quad \Rightarrow { a } _{ 2 }x+{ b } _{ 2 }y={ c } _{ 2 }$
then the solution of $x$ and $y$ can be obtained by evaluating the following integral :
$x=\frac { \left| \underset { { c } _{ 2 } }{ { c } _{ 1 } } \quad \underset { { b } _{ 2 } }{ { b } _{ 1 } }  \right|  }{ \left| \underset { { a } _{ 2 } }{ { a } _{ 1 } } \quad \underset { { b } _{ 2 } }{ { b } _{ 1 } }  \right|  } $  and  $y=\dfrac { \left| \underset { { a } _{ 2 } }{ { a } _{ 1 } } \quad \underset { { c } _{ 2 } }{ { c } _{ 1 } }  \right|  }{ \left| \underset { { a } _{ 2 } }{ { a } _{ 1 } } \quad \underset { { b } _{ 2 } }{ { b } _{ 1 } }  \right|  } $
$\therefore$    $x+y=3$
  $3x-2y=4$
can be solved using the above method
$x=\dfrac { \left| \underset { 4 }{ 3 } \quad \underset { -2 }{ 1 }  \right|  }{ \left| \underset { 3 }{ 1 } \quad \underset { -2 }{ 1 }  \right|  } \quad ;\quad y=\dfrac { \left| \underset { 3 }{ 1 } \quad \underset { 4 }{ 3 }  \right|  }{ \left| \underset { 3 }{ 1 } \quad \underset { -2 }{ 1 }  \right|  } $
$x=\dfrac { -6-4 }{ -2-3 } \quad ;\quad y=\dfrac { 4-9 }{ -5 } $
$x=\dfrac { 10 }{ -5 } \quad ;\quad y=\dfrac { -5 }{ -5 } $
$x=2\quad ;\quad y=1$
now the quantity $\left| \underset { 3 }{ 1 } \quad \underset { -2 }{ 1 }  \right| =D$ (determinant)
$D=\left| \underset { 3 }{ 1 } \quad \underset { -2 }{ 1 }  \right| =-5$
So, answer is option C.
Multiple choice mathematical modelling proof by contradiction similar triangles

The given equation $4xy-x-y=z^2$ has:

  1. three positive integer solutions

  2. one positive integer solutions

  3. two positive integer solutions

  4. no positive integer solutions
Reveal answer Fill a bubble to check yourself
D Correct answer
Explanation
Suppose all the solution of the given equation are positive integers
We write the equation in the equivalent form

$(4x-1)(4y-1)=4z^2+1$.

Let $p$ be a prime divisor of $4x-1$. Then

$4z^2+1\equiv 0$(mod $p$)

or

$(2z)^2\equiv -1$ (mod $p$).

On the other hand, Fermat's theorem yields

$(2z)^{p-1}\equiv 1$ (mod $p$)

hence

$(2z)^{p-1}\equiv (2z^2)^{\frac{p-1}{2}}\equiv (-1)^{\frac{p-1}{2}}\equiv 1$(mod $p$)

This implies that $p \equiv 1$ (mod $4$). It follows that all prime divisors of $4x 1$ are congruent to $1$ modulo $4,$ hence $4x 1 1$ (mod $4$), a contradiction.

Hence they are no positive integer solutions.
Multiple choice maths brackets order operations and algebra using brackets in algebraic expressions order of operations

Solve: $12-[5y+2x(y^2-2x+2)+6y-(y^2-1)]\times 2$.

  1. $8x^2+y^2-4xy^2-8x-22y+10$
  2. $8x^2+2y^2+4xy^2-8x-22y+10$
  3. $8x^2+2y^2-4xy^2-8x-22y+10$
  4. $8x^2+2y^2-4xy^2-8x+22y+10$
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation

$12-[5y+2x(y^2-2x+2)+6y-(y^2-1)]\times 0.1$
We need to follow BODMAS rule.
=> Brackets (parts of a calculation inside brackets always come first).
=> Orders (numbers involving powers or square roots).
=> Division.
=> Multiplication.
=> Addition.
=> Subtraction.
$=$ $12-[5y+2xy^2-4x^2+4x+6y-y^2+1]\times 2$
$=$ $12- [2xy^2-4x^2+4x+11y-y^2+1]\times 2$
$=$ $12-[4xy^2-8x^2+8x+22y-2y^2+2]$
$=$ $12-4xy^2+8x^2-8x-22y+2y^2-2$
$=$ $8x^2+2y^2-4xy^2-8x-22y+10$

Multiple choice maths banking and taxation reading graphs describing different situations using equations to plot lines basics of a straight line

The equation  of a line is given by $3x - 2y = 9$ has how many possible solution?

  1. One solution

  2. No solution

  3. Two solution

  4. Infinitely many solution
Reveal answer Fill a bubble to check yourself
D Correct answer
Explanation

A linear equation in two variables represents a line in the coordinate plane. A line contains infinitely many points, and each point on the line is a valid solution to the equation. Therefore, the equation 3x - 2y = 9 has infinitely many solutions, not just one, two, or none.

Multiple choice applications of quadratic equations solving (simple) problems word problems based on quadratic equations quadratic equation maths

Solve the following equations:
$x^{2} + 2xy + 3xz = 50$,
$2y^{2} + 3yz + yx = 10$,
$3z^{2} + zx + 2zy = 10$.

  1. $x=\pm 4; y=\pm 2; z=\pm 2$
  2. $x=\pm -4; y=\pm -2; z=\pm 2$
  3. $x = \pm 5; y = \pm 1; z = \pm 1$

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

Given equations are ${ x }^{ 2 }+2xy+3xz=50$

$\Rightarrow  x(x+2y+3z)=50$    ........(i),
$ 2{ y }^{ 2 }+3yz+yx=10$
$\Rightarrow  y(2y+3z+x)=10$    ........(ii)
and $ 3{ z }^{ 2 }+zx+2zy=10$
$\Rightarrow  z(3z+x+2y)=10$    ..........(iii)
Dividing (i) by (ii), we get

$\dfrac { x }{ y } =\dfrac { 50 }{ 10 } =5\\ \Rightarrow x=5y$    ......(a)
Dividing (ii) by (iii), we get
$\dfrac { y }{ z } =\dfrac { 10 }{ 10 } =1\\ \Rightarrow z=y$    ..........(b)
Substituting (a) and (b) in (ii)
$y(2y+3y+5y)=10\\ \Rightarrow 10{ y }^{ 2 }=10\\ \Rightarrow y=\pm 1$
From (a), we have
$x=\pm 5$
From (b), we have
$z=\pm 1$

Multiple choice applications of quadratic equations solving (simple) problems word problems based on quadratic equations quadratic equation maths

Solve the following equations:
$x + 2y - z = 11$,
$x^{2} - 4y^{2} + z^{2} = 37$,
$xz = 24$.

  1. $x=2, -5; y=2; z=2,-4$
  2. $x=8,-3; y=3; z=-3,-8$
  3. $x=-3, 5; y=4; z=2,5$
  4. $x=2,4; y=3; z=3, -5$
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

${ x }^{ 2 }-4{ y }^{ 2 }+{ z }^{ 2 }=37$    ......(i)

$xz=24$    .......(ii)
$ x+2y-z=11$    .....(iii)
$\Rightarrow  x-z=11-2y$
On squaring both sides, we have

${ x }^{ 2 }+{ z }^{ 2 }-2xz=121+4{ y }^{ 2 }-44y\\ \Rightarrow { x }^{ 2 }+{ z }^{ 2 }-4{ y }^{ 2 }-2xz=121-44y\\ \Rightarrow 37-2(24)=121-44y\\ \Rightarrow -44y=-132\\ \Rightarrow y=3$
Substituting $y$ in (iii), 
$x+6-z=11\\ \Rightarrow x-z=5$
From (ii), $z=\dfrac { 24 }{ x } $
Thus $x-\dfrac { 24 }{ x } =5$
$ \Rightarrow { x }^{ 2 }-24=5x\\ \Rightarrow { x }^{ 2 }-5x-24=0\\ \Rightarrow { x }^{ 2 }-3x+8x-24=0\\ \Rightarrow x(x-3)+8(x-3)=0\\ \Rightarrow (x+8)(x-3)=0\\ \Rightarrow x=-8,3$
Putting in $ z=\dfrac { 24 }{ x } $
Thus $ z=-3,8$
So, the values of $x$ are $-8,3$, values of $z$ are $-3,8$ and value of $y$ is $3$.

Multiple choice reciprocal equations theory of equations maths

The number of solutions $(x, y, z)$ to the system of equations $ x + 2y + 4z = 9, 4yz + 2xz + xy = 13, xyz = 3 $ such that at least two of $ x, y, z$ are integers is

  1. $3$
  2. $5$
  3. $6$
  4. $4$
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation
Let the roots of the system equation are:
$\alpha =x,\beta =2y,\gamma =4z$
$\alpha +\beta +\gamma =x+2y+4z=9$
$\alpha \beta +\beta \gamma +\gamma \alpha =2xy+8yz+yzx$
$=2(4yz+2xz+xy)\Rightarrow 26$
$\alpha \beta \gamma =8xyz\Rightarrow 24$
Thus,our polynomial should be:
$P^{3}-9P+26P-24=0$
$(P-2)(P-3)(P-4)=0$
since our roots are :
$\alpha =x,\beta =2y$ and $\gamma =4z$
$(x,2y,4z)=(2,3,4)$ or its permutations,or 6 combination.
However,note that one case if,
$x=4,2y=3$ and $4z=2$
$(x,y,z)=(4,\dfrac{3}{2},\dfrac{1}{2})$
which two of the roots are not an integer :Excluding of this case ,we have five solutions.
Multiple choice mathematics and statistics parabola tracing of the parabola definitions related to parabola introduction to parabola

The equation of directrix from the following is,

  1. $2x - y = 0$
  2. $x + 2y = 0$
  3. $x + y = 0$
  4. $x + 3y = 0$
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation
Let $(a,b)$ be the focus and $y=m{x}$ be directrix 

$\implies \bigg(\dfrac{m{x}-y}{\sqrt{1+m^{2}}}\bigg)^{2}=(x-a)^{2}+(y-b)^{2}$

Differentiating on both sides

$\dfrac{(m{x}-y)(m-\dfrac{d{y}}{d{x}})}{1+m^{2}}=(x-a)+(y-b)\dfrac{d{y}}{d{x}}$

$x$ axis is tangent at $(1,0)$

$\dfrac{m(m)}{1+m^{2}}=1-a\implies a=\dfrac{1}{1+m^{2}}$

$y$ axis  is tangent at $(0,2)$

$\dfrac{-2}{1+m^{2}}=b-2\implies b=\dfrac{2{m}^{2}}{1+m^{2}}$

$(1,0)$ lies on parabola

$\dfrac{(m)^{2}}{1+m^{2}}=(1-a)^{2}+b^{2}$

Substituting $a$ and $b$ values 

$\implies (4{m^2}-1)(m^{2})=0\implies m=0,\pm \dfrac{1}{2}$

For $m=0$ we get $a=1,b=0$ which means that the directrix cuts the parabola which is not possible so $m=\pm \dfrac{1}{2}$

$\implies a=\dfrac{4}{5},b=\dfrac{2}{5}$

So the focus is $\bigg(\dfrac{4}{5},\dfrac{2}{5}\bigg)$

the directrix is $2{y}+x=0$

Hence option $B$ is the answer.
Multiple choice graphs to solve linear and non linear equations functions and their graphs curved graphs sets, relations and functions maths

Given, $y=3$, $y=ax^2+b$
In the system of equations above, $a$ and $b$ are constants. For which of the following values of $a$ and $b$ does the system of equations have exactly two real solutions?

  1. $a = 2, b = 2$
  2. $a = 2, b = 4$
  3. $a = 2, b = 3$
  4. $a = 4, b = 3$
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

On substituting value of $y=3$ in second equation, we get
$ax^2+b=3$
$\Rightarrow ax^2+b-3=0$
$\Rightarrow ax^2=3-b$
$\Rightarrow x^2=\frac{3-b}a$
Since $x^2$ is positive quantity, therefore just $a=2$ and $b=2$ satisfies this.
Hence, option A is correct.

Multiple choice graphs to solve linear and non linear equations functions and their graphs curved graphs sets, relations and functions maths

The system of equations:

$\displaystyle y={ x }^{ 2 }-2x$
$\displaystyle y=2x-1$ has two solutions for ($x,y$). 
Determine the greater value of $x$.

  1. $\displaystyle 2-\sqrt { 3 } $
  2. $\displaystyle \sqrt { 3 } $
  3. $\displaystyle 2+2\sqrt { 3 } $
  4. $\displaystyle 5$
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation
Given, $y=x^{2}-2x$
$y=2x-1$
Then $x^{2}-2x=2x-1 $
$\Rightarrow x^{2}-2x-2x+1=0$
$\Rightarrow x^{2}-4x+1=0$
Manipulate this equation $ax^{2}+bx+c=0$
We know $x=\dfrac{-b\pm \sqrt{b^{2}-4ac}}{2a}$
$\therefore x=\dfrac{-(-4)\pm \sqrt{(-4)^{2}-4(1)(1)}}{2(1)}$
$\Rightarrow x= \dfrac{4\pm \sqrt{16-4}}{2}$
$\Rightarrow x=\dfrac{4\pm \sqrt{12}}{2}$
$\Rightarrow x=\dfrac{4\pm 2\sqrt{3}}{2}$
The greater of the two possible values for $x$ is $x=2\pm 2\sqrt{3}$. 
Therefore, the correct answer is (C).
Multiple choice graphs to solve linear and non linear equations functions and their graphs curved graphs sets, relations and functions maths

The set of values of $c$ so that the equations $\displaystyle y=\left | x \right |+c: : and: : x^{2}+y^{2}-8\left | x \right |-9=0 $ have no solution is

  1. $\displaystyle \left ( -\infty ,-3 \right )\cup \left ( 3,\infty \right )$
  2. $(-3, 3)$
  3. $\displaystyle \left ( -\infty ,-5\sqrt{2} \right )\cup \left ( 5\sqrt{2},\infty \right )$
  4. $\displaystyle \left ( -\infty ,-4-5\sqrt{2} \right )\cup \left ( 5\sqrt{2}-4,\infty \right )$
Reveal answer Fill a bubble to check yourself
D Correct answer
Explanation

Given equation 

$y=\left |x\right |+c$---(1)
$x^2+y^2-8\left| x\right |-9=0$
From equation (1) and (2)
$x^2+(\left | x\right |+c)^2-8\left |x\right |-9=0$
when $x>0$
$x^2+(x+c)^2-8x-9=0$
$x^2+x^2+c^2+2cx-8x-9=0$
$2x^2+x(2c-8)+c^2-9=0$
For no solution 
$D<0$
$(2c-8)^2-4\times 2 (c^2-9)<0$
$4c^2+64-32c-8c^2+72<0$
$-4c^2-32c+136<0$
$c^2+8c-34>0$
$c=\dfrac{-8\pm\sqrt{64+136}}{2}$

$c=\dfrac{-8\pm\sqrt{200}}{2}$

$c=-4\pm 5\sqrt{2}$
C has root $c=-4\pm5\sqrt{2}$
Hence for no solution c has all value excluding it's roots  
$c\epsilon(-\infty,-4-5\sqrt{2})\cup(5\sqrt{2}-4,\infty)$

Multiple choice maths solving equations numerically finding roots by iteration fundamental theorem of algebra complex numbers and linear inequations

Solve the simultaneous equations using the convergent iterations:
$5x$ + $y$ + $2z$ = $19$
$2x$ + $3y$ +$8z$ = $39$
$x$ + $4y$ -$2z$ = $-2$

  1. $x=2$, $y=2$ and $\text z =1$
  2. $x=2$, $y=1$ and $\text z =4$
  3. $x=3$, $y=2$ and $\text z =2$

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

$5x+y+2z=19$

$x+4y-2z=-2$

Adding above two equations we get,
$5x+y+2z+x+4y-2z=19-2$
$\implies 6x+5y=17$

Taking another equation 
$2x+3y+8z=39$

Multiplying the given last equation with $4$
$4x+16y-8z=-8$

Adding above two equations 
$2x+3y+8z+4x+16y-8z=31$
$\implies 6x+19y=31$

Now,
$6x+19y-(6x+5y)=31-17$
$\implies 19y-5y=14$
$\implies y=1$

By substituting the value of $y$ into the above equations 
we get,
$x=2$ and $z=4$
$\therefore\ x=2,y=1$ and $z=4$.