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

Solve the following system of equations: 2x + 3y = 7 4x - 2y = 10

  1. (x, y) = (2, 1)

  2. (x, y) = (1, 2)

  3. (x, y) = (3, -1)

  4. (x, y) = (-1, 3)
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

To solve this system of equations, we can use the elimination method. Multiplying the first equation by 2 and the second equation by 3, we get: 4x + 6y = 14 12x - 6y = 30 Adding the two equations, we get: 16x = 44 Solving for x, we get x = 2. Substituting x = 2 back into either of the original equations, we can solve for y. Using the first equation, we get 2(2) + 3y = 7, which gives y = 1. Therefore, the solution is (x, y) = (2, 1).

Multiple choice

Solve the following system of equations: y = x^2 - 1 y = 2x - 3

  1. (x, y) = (2, 1)

  2. (x, y) = (1, 2)

  3. (x, y) = (3, -1)

  4. (x, y) = (-1, 3)
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

To solve this system of equations, we can substitute the first equation into the second equation. Substituting y = x^2 - 1 into y = 2x - 3, we get x^2 - 1 = 2x - 3. Rearranging the equation, we get x^2 - 2x + 2 = 0. Factoring the quadratic equation, we get (x - 2)(x - 1) = 0. Setting each factor equal to zero, we get x = 2 and x = 1. Substituting x = 2 back into the first equation, we get y = 2^2 - 1 = 3. Therefore, the solution is (x, y) = (2, 1).

Multiple choice

Solve the following system of equations: 3x + 2y = 5 2x - y = 1

  1. (x, y) = (1, 2)

  2. (x, y) = (2, 1)

  3. (x, y) = (3, -1)

  4. (x, y) = (-1, 3)
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

To solve this system of equations, we can use the substitution method. Solving the second equation for y, we get y = 2x - 1. Substituting this into the first equation, we get 3x + 2(2x - 1) = 5. Solving for x, we get x = 1. Substituting x = 1 back into the second equation, we get y = 2(1) - 1 = 1. Therefore, the solution is (x, y) = (1, 2).

Multiple choice

Solve the following system of equations: 2x + 3y = 7 4x - 2y = 10

  1. (x, y) = (2, 1)

  2. (x, y) = (1, 2)

  3. (x, y) = (3, -1)

  4. (x, y) = (-1, 3)
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

To solve this system of equations, we can use the elimination method. Multiplying the first equation by 2 and the second equation by 3, we get: 4x + 6y = 14 12x - 6y = 30 Adding the two equations, we get: 16x = 44 Solving for x, we get x = 2. Substituting x = 2 back into either of the original equations, we can solve for y. Using the first equation, we get 2(2) + 3y = 7, which gives y = 1. Therefore, the solution is (x, y) = (2, 1).

Multiple choice

Solve the following system of equations: y = x^2 - 1 y = 2x - 3

  1. (x, y) = (2, 1)

  2. (x, y) = (1, 2)

  3. (x, y) = (3, -1)

  4. (x, y) = (-1, 3)
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

To solve this system of equations, we can substitute the first equation into the second equation. Substituting y = x^2 - 1 into y = 2x - 3, we get x^2 - 1 = 2x - 3. Rearranging the equation, we get x^2 - 2x + 2 = 0. Factoring the quadratic equation, we get (x - 2)(x - 1) = 0. Setting each factor equal to zero, we get x = 2 and x = 1. Substituting x = 2 back into the first equation, we get y = 2^2 - 1 = 3. Therefore, the solution is (x, y) = (2, 1).

Multiple choice

Solve the following system of equations: 3x + 2y = 5 2x - y = 1

  1. (x, y) = (1, 2)

  2. (x, y) = (2, 1)

  3. (x, y) = (3, -1)

  4. (x, y) = (-1, 3)
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

To solve this system of equations, we can use the substitution method. Solving the second equation for y, we get y = 2x - 1. Substituting this into the first equation, we get 3x + 2(2x - 1) = 5. Solving for x, we get x = 1. Substituting x = 1 back into the second equation, we get y = 2(1) - 1 = 1. Therefore, the solution is (x, y) = (1, 2).

Multiple choice

Solve the following system of equations: 2x + 3y = 7 4x - 2y = 10

  1. (x, y) = (2, 1)

  2. (x, y) = (1, 2)

  3. (x, y) = (3, -1)

  4. (x, y) = (-1, 3)
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

To solve this system of equations, we can use the elimination method. Multiplying the first equation by 2 and the second equation by 3, we get: 4x + 6y = 14 12x - 6y = 30 Adding the two equations, we get: 16x = 44 Solving for x, we get x = 2. Substituting x = 2 back into either of the original equations, we can solve for y. Using the first equation, we get 2(2) + 3y = 7, which gives y = 1. Therefore, the solution is (x, y) = (2, 1).

Multiple choice

Solve the following system of equations: y = x^2 - 1 y = 2x - 3

  1. (x, y) = (2, 1)

  2. (x, y) = (1, 2)

  3. (x, y) = (3, -1)

  4. (x, y) = (-1, 3)
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

To solve this system of equations, we can substitute the first equation into the second equation. Substituting y = x^2 - 1 into y = 2x - 3, we get x^2 - 1 = 2x - 3. Rearranging the equation, we get x^2 - 2x + 2 = 0. Factoring the quadratic equation, we get (x - 2)(x - 1) = 0. Setting each factor equal to zero, we get x = 2 and x = 1. Substituting x = 2 back into the first equation, we get y = 2^2 - 1 = 3. Therefore, the solution is (x, y) = (2, 1).

Multiple choice

Solve the following system of equations: 3x + 2y = 5 2x - y = 1

  1. (x, y) = (1, 2)

  2. (x, y) = (2, 1)

  3. (x, y) = (3, -1)

  4. (x, y) = (-1, 3)
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

To solve this system of equations, we can use the substitution method. Solving the second equation for y, we get y = 2x - 1. Substituting this into the first equation, we get 3x + 2(2x - 1) = 5. Solving for x, we get x = 1. Substituting x = 1 back into the second equation, we get y = 2(1) - 1 = 1. Therefore, the solution is (x, y) = (1, 2).

Multiple choice

Solve the following system of equations: 2x + 3y = 7 4x - 2y = 10

  1. (x, y) = (2, 1)

  2. (x, y) = (1, 2)

  3. (x, y) = (3, -1)

  4. (x, y) = (-1, 3)
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

To solve this system of equations, we can use the elimination method. Multiplying the first equation by 2 and the second equation by 3, we get: 4x + 6y = 14 12x - 6y = 30 Adding the two equations, we get: 16x = 44 Solving for x, we get x = 2. Substituting x = 2 back into either of the original equations, we can solve for y. Using the first equation, we get 2(2) + 3y = 7, which gives y = 1. Therefore, the solution is (x, y) = (2, 1).

Multiple choice

Which of the following is an example of an indeterminate equation?

  1. $x + y = 5$
  2. $x^2 + y^2 = 1$
  3. $x^3 + y^3 = z^3$
  4. $x^2 - y^2 = 1$
Reveal answer Fill a bubble to check yourself
D Correct answer
Explanation

An indeterminate equation is one that has infinitely many solutions. $x^2 - y^2 = 1$ is an example of an indeterminate equation, as it has infinitely many integer solutions for $x$ and $y$.

Multiple choice

What is the solution to the equation x^2 - 4x + 3 = 0?

  1. x = 1, x = 3

  2. x = 2, x = 4

  3. x = 3, x = 5

  4. x = 4, x = 6
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

We can solve this equation using the quadratic formula: x = (-b ± √(b^2 - 4ac)) / 2a. Substituting a = 1, b = -4, and c = 3, we get x = (-(-4) ± √((-4)^2 - 4(1)(3))) / 2(1) = (4 ± √(16 - 12)) / 2 = (4 ± √4) / 2 = (4 ± 2) / 2. Therefore, the solutions are x = 1 and x = 3.

Multiple choice

Solve the system of equations: (x + 2y = 5) and (2x - y = 1).

  1. \((x, y) = (1, 2)\)
  2. \((x, y) = (2, 1)\)
  3. \((x, y) = (3, 0)\)
  4. \((x, y) = (0, 3)\)
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

To solve the system of equations, we can use the substitution method. First, solve one of the equations for one of the variables. For example, we can solve the first equation for (x): (x = 5 - 2y). Then, substitute this expression for (x) into the other equation: (2(5 - 2y) - y = 1). This gives us (10 - 4y - y = 1), which simplifies to (-5y = -9). Dividing both sides by (-5), we get (y = 9/5). Substituting this value of (y) back into the first equation, we get (x + 2(9/5) = 5), which simplifies to (x = 1). Therefore, the solution to the system of equations is ((x, y) = (1, 2)).

Multiple choice

Solve the system of equations: (3x + 2y = 7) and (2x - y = 1).

  1. \((x, y) = (1, 2)\)
  2. \((x, y) = (2, 3)\)
  3. \((x, y) = (3, 4)\)
  4. \((x, y) = (4, 5)\)
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

To solve the system of equations, we can use the elimination method. First, multiply the second equation by 2 to get (4x - 2y = 2). Then, add this equation to the first equation to get (7x = 9). Dividing both sides by 7, we get (x = 9/7). Substituting this value of (x) back into the first equation, we get (3(9/7) + 2y = 7), which simplifies to (y = 3). Therefore, the solution to the system of equations is ((x, y) = (2, 3)).

Multiple choice

Solve the system of equations: (x + y = 5) and (x - y = 1).

  1. \((x, y) = (2, 3)\)
  2. \((x, y) = (3, 2)\)
  3. \((x, y) = (4, 1)\)
  4. \((x, y) = (1, 4)\)
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

To solve the system of equations, we can use the addition method. First, add the two equations together to get (2x = 6). Dividing both sides by 2, we get (x = 3). Substituting this value of (x) back into the first equation, we get (3 + y = 5), which simplifies to (y = 2). Therefore, the solution to the system of equations is ((x, y) = (3, 2)).