Mathematics · Quantitative Aptitude

Linear Equations

153 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

Which of the following equations is a linear equation?

  1. 3x + 2yx2 = 4

  2. 2x + 3y2 = 4

  3. 2x + 2y + x2 = 7

  4. x = 4
Reveal answer Fill a bubble to check yourself
D Correct answer
Explanation

The highest power of the term in the given equation is called degree. An equation having degree one, is called a linear equation. $\because$x = 4 is an equation of degree one, therefore it is a linear equation.

Multiple choice

Which of the following equations is a linear equation?

  1. 7x + 10 = 0

  2. 2x + 3yz = 0

  3. 7xy + 2y = 0

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

The highest power of the term in the given equation is called degree. The equation having a degree one is called a linear equation. Therefore, 7x + 10 = 0 is the linear equation.

Multiple choice

Which of the following equations is a linear equation?

  1. xy + 2x2 = 3

  2. x + y - 3 = 0

  3. x2 + y2 - 3xy = 4

  4. x2 - y2 = 5
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

The highest power of the term in the given equation is called degree. The equation having a degree one is called a linear equation. Therefore, the equation x + y - 3 = 0 is a linear equation.

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

$x + 2y + 3z = 0$ 

$2x + 3y + 4z = 0$
$7x + 13y + 19z =0$
The system of equations have non trivial solutions

  1. True

  2. False
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

A system of homogeneous linear equations has non-trivial solutions if the determinant of the coefficient matrix is zero. The determinant of [[1, 2, 3], [2, 3, 4], [7, 13, 19]] is 1(57-52) - 2(38-28) + 3(26-21) = 5 - 20 + 15 = 0. Since the determinant is zero, non-trivial solutions exist.

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

$2x - 3y + z = 0$ 

$x + 2y - 3z =0$
$4x - y - 2z = 0$
The system of equations have a non trivial solution

  1. True

  2. False
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

For the system to have non-trivial solutions, the determinant must be zero. The determinant of [[2, -3, 1], [1, 2, -3], [4, -1, -2]] is 2(-4-3) - (-3)(-2+12) + 1(-1-8) = 2(-7) + 3(10) - 9 = -14 + 30 - 9 = 7. Since the determinant is not zero, the system has only the trivial solution (0,0,0).

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

Which of the following are correct in respect of the system of equations $x + y + z = 8, x - y + 2z = 6$ and $3x - y + 5z = k$?
1. They have no solution, if $k = 15$.
2. They have infinitely many solutions, if $k = 20$.
3. They have unique solution, if $k = 25$.
Select the correct answer using the code given below

  1. 1 and 2 only

  2. 2 and 3 only

  3. 1 and 3 only

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

First we need to know if the above equations are linearly dependent or no,

In order to figure that out lets find the determinant $D$ which is;
$D=\left| \begin{matrix} 1 & 1 & 1 \ 1 & -1 & 2 \ 3 & -1 & 5 \end{matrix} \right| =0$
as $D=0$ the equations are linearly dependent, which means the system of equations can either be inconsistent or have infinitely many solutions. That shall depend on the value of k.
For there to be inifinitely many solutions , values of ${ D } _{ 1 }, { D } _{ 2 }, { D } _{ 3 }$ should be $0$ , defined as;
${ D } _{ 1 }=\left| \begin{matrix} 8 & 1 & 1 \ 6 & -1 & 2 \ k & -1 & 5 \end{matrix} \right| \ { D } _{ 2 }=\left| \begin{matrix} 1 & 8 & 1 \ 1 & 6 & 2 \ 3 & k & 5 \end{matrix} \right| \ { D } _{ 3 }=\left| \begin{matrix} 1 & 1 & 8 \ 1 & -1 & 6 \ 3 & -1 & k \end{matrix} \right| $
which on solving you get;
${ D } _{ 1 }=\left| \begin{matrix} 8 & 1 & 1 \ 6 & -1 & 2 \ k & -1 & 5 \end{matrix} \right| =3k-60=3(k-20)\ { D } _{ 2 }=\left| \begin{matrix} 1 & 8 & 1 \ 1 & 6 & 2 \ 3 & k & 5 \end{matrix} \right| =k-20\ { D } _{ 3 }=\left| \begin{matrix} 1 & 1 & 8 \ 1 & -1 & 6 \ 3 & -1 & k \end{matrix} \right| =40-2k=-2(k-20)$
Now for all the Determinants to be zero , its evident that $k=20$
Hence statement 2 is correct.
Subsequently for there to be no solution $k\neq 20$, hence statement 1 is correct as well,
As $D$ is zero the system of equations CANNOT have a unique solution hence, statement 3 is WRONG.

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 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.