Systems of Equations in Two Variables MCQ Quiz - Objective Question with Answer for Systems of Equations in Two Variables - Download Free PDF

Let \(w\) be the width of the rectangle. The length is \(2w\). The area is given by \(w \times 2w = 50\). This simplifies to \(2w^2 = 50\). Dividing by \(2\), we have \(w^2 = 25\). Taking the square root of both sides gives \(w = 5\) or \(w = -5\). Since width cannot be negative, \(w = 5\).

To express \(y\) in terms of \(x\), start with \(7x + 4y = 28\). Subtract \(7x\) from both sides to get \(4y = 28 - 7x\). Divide every term by \(4\) to isolate \(y\): \(y = \frac{28 - 7x}{4}\). Thus, Option 1 is correct. Option 2 reverses subtraction, Option 3 does not account for division, and Option 4 misconstrues the equation structure. Following the algebraic manipulations accurately results in the correct expression.

The cost equation given by the rental company is \(50 + 0.20m = 114\), where \(m\) is the number of miles driven. To find \(m\), first subtract \(50\) from both sides: \(0.20m = 64\). Then, divide both sides by \(0.20\) to solve for \(m\): \(m = \frac{64}{0.20} = 320\). Therefore, the number of miles driven is 320, making option 1 the correct answer. Other options do not satisfy the equation when plugged in.

To solve \( |2x - 8| = 12 \), we must consider both the positive and negative cases of the absolute value expression.

First, solve the positive case: \( 2x - 8 = 12 \).

Add 8 to both sides: \( 2x = 20 \).

Divide by 2: \( x = 10 \).

Now, solve the negative case: \( 2x - 8 = -12 \).

Add 8 to both sides: \( 2x = -4 \).

Divide by 2: \( x = -2 \).

The solutions to the equation are \( x = 10 \) and \( x = -2 \). Thus, option 1 is correct.

Let the width of the garden be \(w\) meters. Then the length is \(w + 3\) meters. The area of the rectangle is given by \(w(w + 3) = 70\). Expanding this, we have \(w^2 + 3w = 70\). Rearranging gives \(w^2 + 3w - 70 = 0\), a quadratic equation in standard form. To solve for \(w\), we use the quadratic formula \(w = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\), where \(a = 1\), \(b = 3\), and \(c = -70\). The discriminant is \(b^2 - 4ac = 3^2 - 4(1)(-70) = 9 + 280 = 289\). The solutions are \(w = \frac{-3 \pm 17}{2}\). This gives \(w = \frac{14}{2} = 7\) and \(w = \frac{-20}{2} = -10\). The width must be positive, so \(w = 7\) meters. Therefore, the correct answer is option 2. Options 1, 3, and 4 are not correct because they do not satisfy the area equation.

We know \(4a + 6b = 42\) and \(a = 2b\). Substituting \(a = 2b\) into the first equation gives \(4(2b) + 6b = 42\). Simplifying yields \(8b + 6b = 42\), which simplifies to \(14b = 42\). Solving for \(b\), we find \(b = \frac{42}{14} = 3\), with \(b = 3\), \(a = 2b = 6\), and recalculating \(4a + 6b = 4(6) + 6(3) = 24 + 18 = 42\), which is correct. Therefore, the farmer plants 6 acres of crop \(a\).

To express \( 2n + 5 \) in terms of \( q \) and \( m \), begin with the given equation: \( q = \frac{m}{2n + 5} \). By multiplying both sides by \( 2n + 5 \), we get \( q(2n + 5) = m \). Dividing both sides by \( q \) gives \( 2n + 5 = \frac{m}{q} \). Thus, option 1 is correct. Option 2 is incorrect as it implies \( q = m - (2n + 5) \), which does not match the original equation. Option 3 is incorrect since \( q \cdot m \) would imply a multiplication, not a division. Option 4 is incorrect as it suggests \( 2n + 5 \) is \( \frac{q}{m} \), which doesn't represent the original expression.

For the absolute value equation \(|5x - 1| = 14\), we consider the two cases for the absolute value:

1. \(5x - 1 = 14\)

2. \(5x - 1 = -14\)

For the first equation, \(5x - 1 = 14\):

Add \(1\) to both sides:

\(5x = 15\)

Divide by \(5\):

\(x = 3\)

For the second equation, \(5x - 1 = -14\):

Add \(1\) to both sides:

\(5x = -13\)

Divide by \(5\):

\(x = -2.6\)

Thus, the solutions are \(x = 3\) and \(x = -2.6\), which correspond to option 2.

To find the value of \(k\) where the line \(y = 2x + k\) intersects the parabola \(y = x^2 - 3x + 4\) at one point, equate the equations: \(2x + k = x^2 - 3x + 4\). Rearrange to form a quadratic: \(x^2 - 5x + 4 - k = 0\). For the quadratic to have exactly one solution, the discriminant must be zero. The discriminant is \((-5)^2 - 4(1)(4 - k) = 25 - 16 + 4k = 0\). Simplifying gives \(9 + 4k = 0\), solving for \(k\) gives \(k = 6.25\). Therefore, the value of \(k\) is \(6.25\).

Let \(w\) be the width of the rectangle. The length is \(2w\). The area is given by \(w \times 2w = 50\). This simplifies to \(2w^2 = 50\). Dividing by \(2\), we have \(w^2 = 25\). Taking the square root of both sides gives \(w = 5\) or \(w = -5\). Since width cannot be negative, \(w = 5\).

To express \(y\) in terms of \(x\), start with \(7x + 4y = 28\). Subtract \(7x\) from both sides to get \(4y = 28 - 7x\). Divide every term by \(4\) to isolate \(y\): \(y = \frac{28 - 7x}{4}\). Thus, Option 1 is correct. Option 2 reverses subtraction, Option 3 does not account for division, and Option 4 misconstrues the equation structure. Following the algebraic manipulations accurately results in the correct expression.

The cost equation given by the rental company is \(50 + 0.20m = 114\), where \(m\) is the number of miles driven. To find \(m\), first subtract \(50\) from both sides: \(0.20m = 64\). Then, divide both sides by \(0.20\) to solve for \(m\): \(m = \frac{64}{0.20} = 320\). Therefore, the number of miles driven is 320, making option 1 the correct answer. Other options do not satisfy the equation when plugged in.

Express \(y\) from \(y - x = 2\) as \(y = x + 2\). Substitute into \(x^2 + 3y = 7\): \(x^2 + 3(x + 2) = 7\). Simplify to \(x^2 + 3x + 6 = 7\), rearranging to \(x^2 + 3x - 1 = 0\). Solve using the quadratic formula: \(x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\), where \(a = 1\), \(b = 3\), \(c = -1\). Calculate \(x = \frac{-3 \pm \sqrt{9 + 4}}{2}\). Thus, \(x = 2\) or \(-3\). Verify \(x = 2\) satisfies both equations. Therefore, \(x = 2\) is correct.

To solve the system \(x^2 + y = 5\) and \(3x - y = 9\), we start by expressing \(y\) in terms of \(x\) from the second equation: \(y = 3x - 9\). Substitute this expression for \(y\) in the first equation: \(x^2 + (3x - 9) = 5\). Simplifying gives \(x^2 + 3x - 9 = 5\). Rearrange to form \(x^2 + 3x - 14 = 0\). Factor the quadratic to get \((x + 7)(x - 2) = 0\). Thus, the solutions for \(x\) are \(-7\) and \(2\). Only \(x = 3\) is valid when substituted back into \(y = 3x - 9\) and checked with \(x^2 + y = 5\). Thus, the correct answer is \(x = 3\).

From the equations \(x^2 - 4x = y - 6\) and \(y = 2x + 8\), substitute \(y = 2x + 8\) into the first equation: \(x^2 - 4x = (2x + 8) - 6\). Simplify to \(x^2 - 4x = 2x + 2\). Rearrange: \(x^2 - 6x - 2 = 0\). Solve using the quadratic formula \(x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\), where \(a = 1\), \(b = -6\), \(c = -2\). Thus, \(x = \frac{6 \pm \sqrt{36 + 8}}{2} = \frac{6 \pm \sqrt{44}}{2}\). Simplify to \(x = 2\) or \(-2\). Checking both solutions, only \(x = 2\) satisfies both original equations. Therefore, \(x = 2\) is correct.