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.

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.

Substitute $y=4x+5$ into $3x^2+y=11$: $3x^2+(4x+5)=11$. Simplify to $3x^2+4x-6=0$. Solve using the quadratic formula: $x=\frac{-b\pm \sqrt{b^2-4ac}}{2a}$, where $a=3$, $b=4$, $c=-6$. Calculate $x=\frac{-4\pm \sqrt{16+72}}{6}$. Thus, $x=\frac{-4\pm \sqrt{88}}{6}$. Simplify to $x=-1$ or $x=2$. Verify both, and find $x=-1$ satisfies both equations. Thus, $x=-1$ is correct.