Which of the following uses the substitution method to solve linear-quadratic system y=x2−4x+8 , y=3x−4 ?(1 point)

Answer 1

To solve the system of equations using the substitution method, let's set the equations equal to each other. We have the two equations:

1. \( y = x^2 - 4x + 8 \)
2. \( y = 3x - 4 \)

Since both equations are set equal to \( y \), we can substitute the expression for \( y \) from the second equation into the first equation:

\[ 3x - 4 = x^2 - 4x + 8 \]

Now, we rearrange the equation to set it to zero:

\[ x^2 - 4x - 3x + 8 + 4 = 0 \] \[ x^2 - 7x + 12 = 0 \]

Next, we can factor the quadratic equation:

\[ (x - 3)(x - 4) = 0 \]

Setting each factor to zero gives us the solutions for \( x \):

1. \( x - 3 = 0 \) → \( x = 3 \)
2. \( x - 4 = 0 \) → \( x = 4 \)

Now, we can substitute these values back into one of the original equations to find the corresponding \( y \) values. Using the linear equation \( y = 3x - 4 \):

For \( x = 3 \): \[ y = 3(3) - 4 = 9 - 4 = 5 \]

For \( x = 4 \): \[ y = 3(4) - 4 = 12 - 4 = 8 \]

Thus, the solutions to the system are:

1. \( (3, 5) \)
2. \( (4, 8) \)

So the system of equations is solved using the substitution method by equating the two expressions for \( y \) and then solving for \( x \).

Answer 2

Use the substitution method to find the solutions to the linear-quadratic system y=x2−x−40, y=−3x−5?(1 point) Responses

Answer 3

To solve the system of equations using the substitution method, we start with the given equations:

1. \( y = x^2 - x - 40 \)
2. \( y = -3x - 5 \)

Since both equations are equal to \( y \), we can set them equal to each other:

\[ x^2 - x - 40 = -3x - 5 \]

Next, we will rearrange the equation to form a standard quadratic equation by bringing all terms to one side:

\[ x^2 - x + 3x - 40 + 5 = 0 \]

This simplifies to:

\[ x^2 + 2x - 35 = 0 \]

Now, we can factor this quadratic equation:

\[ (x + 7)(x - 5) = 0 \]

Setting each factor to zero gives us the solutions for \( x \):

1. \( x + 7 = 0 \) → \( x = -7 \)
2. \( x - 5 = 0 \) → \( x = 5 \)

Now, we can substitute these values of \( x \) back into one of the original equations to find the corresponding \( y \) values. We'll use the linear equation \( y = -3x - 5 \):

For \( x = -7 \): \[ y = -3(-7) - 5 = 21 - 5 = 16 \]

For \( x = 5 \): \[ y = -3(5) - 5 = -15 - 5 = -20 \]

Thus, the solutions to the system are:

1. \( (-7, 16) \)
2. \( (5, -20) \)

So the pairs \( (-7, 16) \) and \( (5, -20) \) are the solutions to the given linear-quadratic system.

Answer 4

Use the elimination method to solve the linear-quadratic system y=x2+10 , y=−7x−2 .(1 point) Responses

Answer 5

To solve the system of equations using the elimination method, we will first write down the two equations:

1. \( y = x^2 + 10 \)
2. \( y = -7x - 2 \)

Since both equations equal \( y \), we can set them equal to each other:

\[ x^2 + 10 = -7x - 2 \]

Now, we will rearrange the equation to bring all terms to one side:

\[ x^2 + 7x + 10 + 2 = 0 \]

This simplifies to:

\[ x^2 + 7x + 12 = 0 \]

Next, we can factor the quadratic equation:

\[ (x + 3)(x + 4) = 0 \]

Setting each factor to zero gives us the solutions for \( x \):

1. \( x + 3 = 0 \) → \( x = -3 \)
2. \( x + 4 = 0 \) → \( x = -4 \)

Now, we will substitute these values of \( x \) back into one of the original equations to find the corresponding \( y \) values. We can use either equation; let’s use \( y = -7x - 2 \):

For \( x = -3 \): \[ y = -7(-3) - 2 = 21 - 2 = 19 \]

For \( x = -4 \): \[ y = -7(-4) - 2 = 28 - 2 = 26 \]

Thus, the solutions to the system are:

1. \( (-3, 19) \)
2. \( (-4, 26) \)

So the pairs \( (-3, 19) \) and \( (-4, 26) \) are the solutions to the given linear-quadratic system.

Answer 6

Determine the number of reasonable solutions there can be in a system of equations containing one linear equation and one quadratic equation.(1 point)

Answer 7

In a system of equations consisting of one linear equation and one quadratic equation, there can be up to two reasonable solutions.

Here's a brief explanation of why this is the case:

1. Linear Equation: A linear equation represents a straight line. It can intersect various shapes depending on their nature.

2. Quadratic Equation: A quadratic equation typically represents a parabola. The parabola can open upwards or downwards.

When you graph a linear equation and a quadratic equation together, the following scenarios can occur:

• No intersections: The line is above or below the parabola at all points, yielding no solutions.
• One intersection: The line may just touch the parabola (tangentially) at one point, yielding one solution.
• Two intersections: The line can cut through the parabola at two distinct points, yielding two solutions.

Therefore, a system with one linear equation and one quadratic equation can have 0, 1, or 2 reasonable solutions, but the most it can have is two.

Answer 8

Determine the solution(s) to the following system. y=−x2 y=12x−3

Answer 9

To find the solution(s) to the system of equations given by

1. \( y = -x^2 \)
2. \( y = 12x - 3 \)

we can set the equations equal to each other, since both are equal to \( y \):

\[ -x^2 = 12x - 3. \]

Next, we'll rearrange the equation to set it to zero:

\[ -x^2 - 12x + 3 = 0 \]

Multiplying through by -1 to make it more standard gives us:

\[ x^2 + 12x - 3 = 0. \]

Now we can solve the quadratic equation using the quadratic formula:

\[ x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \]

In this case, \( a = 1 \), \( b = 12 \), and \( c = -3 \). Plugging in the values:

\[ x = \frac{-12 \pm \sqrt{12^2 - 4 \cdot 1 \cdot (-3)}}{2 \cdot 1} \] \[ x = \frac{-12 \pm \sqrt{144 + 12}}{2} \] \[ x = \frac{-12 \pm \sqrt{156}}{2} \] \[ x = \frac{-12 \pm 2\sqrt{39}}{2} \] \[ x = -6 \pm \sqrt{39}. \]

This gives us the two possible solutions for \( x \):

1. \( x = -6 + \sqrt{39} \)
2. \( x = -6 - \sqrt{39} \)

Next, we substitute these \( x \)-values back into one of the original equations to find the corresponding \( y \)-values. Using \( y = 12x - 3 \):

For \( x = -6 + \sqrt{39} \): \[ y = 12(-6 + \sqrt{39}) - 3 = -72 + 12\sqrt{39} - 3 = -75 + 12\sqrt{39} \]

For \( x = -6 - \sqrt{39} \): \[ y = 12(-6 - \sqrt{39}) - 3 = -72 - 12\sqrt{39} - 3 = -75 - 12\sqrt{39} \]

Thus, the solutions to the system are:

1. \( \left(-6 + \sqrt{39}, -75 + 12\sqrt{39}\right) \)
2. \( \left(-6 - \sqrt{39}, -75 - 12\sqrt{39}\right) \)

These represent the two points of intersection between the linear and quadratic functions.