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

Let be the width of the rectangle. The length is . The area is given by . This simplifies to . Dividing by , we have . Taking the square root of both sides gives or . Since width cannot be negative, .

To express in terms of , start with . Subtract from both sides to get . Divide every term by to isolate : . 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 , where is the number of miles driven. To find , first subtract from both sides: . Then, divide both sides by to solve for : . 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 , we must consider both the positive and negative cases of the absolute value expression.

First, solve the positive case: .

Add 8 to both sides: .

Divide by 2: .

Now, solve the negative case: .

Add 8 to both sides: .

Divide by 2: .

The solutions to the equation are and . Thus, option 1 is correct.

Let the width of the garden be meters. Then the length is meters. The area of the rectangle is given by . Expanding this, we have . Rearranging gives , a quadratic equation in standard form. To solve for , we use the quadratic formula , where , , and . The discriminant is . The solutions are . This gives and . The width must be positive, so 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 and . Substituting into the first equation gives . Simplifying yields , which simplifies to . Solving for , we find , with , , and recalculating , which is correct. Therefore, the farmer plants 6 acres of crop .

To express in terms of and , begin with the given equation: . By multiplying both sides by , we get . Dividing both sides by gives . Thus, option 1 is correct. Option 2 is incorrect as it implies , which does not match the original equation. Option 3 is incorrect since would imply a multiplication, not a division. Option 4 is incorrect as it suggests is , which doesn't represent the original expression.

For the absolute value equation , we consider the two cases for the absolute value:

1.

2.

For the first equation, :

Add to both sides:

Divide by :

For the second equation, :

Add to both sides:

Divide by :

Thus, the solutions are and , which correspond to option 2.

To find the value of where the line intersects the parabola at one point, equate the equations: . Rearrange to form a quadratic: . For the quadratic to have exactly one solution, the discriminant must be zero. The discriminant is . Simplifying gives , solving for gives . Therefore, the value of is .

Let be the width of the rectangle. The length is . The area is given by . This simplifies to . Dividing by , we have . Taking the square root of both sides gives or . Since width cannot be negative, .

To express in terms of , start with . Subtract from both sides to get . Divide every term by to isolate : . 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 , where is the number of miles driven. To find , first subtract from both sides: . Then, divide both sides by to solve for : . 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 from as . Substitute into : . Simplify to , rearranging to . Solve using the quadratic formula: , where , , . Calculate . Thus, or . Verify satisfies both equations. Therefore, is correct.

To solve the system and , we start by expressing in terms of from the second equation: . Substitute this expression for in the first equation: . Simplifying gives . Rearrange to form . Factor the quadratic to get . Thus, the solutions for are and . Only is valid when substituted back into and checked with . Thus, the correct answer is .

From the equations and , substitute into the first equation: . Simplify to . Rearrange: . Solve using the quadratic formula , where , , . Thus, . Simplify to or . Checking both solutions, only satisfies both original equations. Therefore, is correct.