Solved on Jan 22, 2024

Find the points of intersection of the given pairs of linear and quadratic equations: $y=x^{2}+4 x+3, y=5 x+9$ and $y=-x^{2}-4 x+6, y=x-8$ .

STEP 1

Assumptions
1. We are given two pairs of equations, each pair representing two lines in the Cartesian plane.
2. We are to find the points of intersection for each pair of equations.
3. The solutions to the equations are the x and y coordinates where the two lines intersect.
4. We will solve each pair of equations separately.

STEP 2

For the first pair of equations, we have:
$y = x^{2} + 4x + 3$ $y = 5x + 9$
To find the points of intersection, we will set the two equations equal to each other, since at the point of intersection, y will have the same value in both equations.

STEP 3

Set the two expressions for y equal to each other:
$x^{2} + 4x + 3 = 5x + 9$

STEP 4

Subtract $5x + 9$ from both sides to bring all terms to one side of the equation and set it equal to zero:
$x^{2} + 4x + 3 - 5x - 9 = 0$

STEP 5

Combine like terms:
$x^{2} - x - 6 = 0$

STEP 6

Factor the quadratic equation:
$(x - 3)(x + 2) = 0$

STEP 7

Set each factor equal to zero and solve for x:
$x - 3 = 0 \quad \text{or} \quad x + 2 = 0$

STEP 8

Solve each equation for x:
$x = 3 \quad \text{or} \quad x = -2$

STEP 9

Now that we have the x-values of the points of intersection, we need to find the corresponding y-values. We can do this by plugging the x-values into either of the original equations. We will use $y = 5x + 9$ .

STEP 10

Plug $x = 3$ into $y = 5x + 9$ :
$y = 5(3) + 9$

STEP 11

Calculate the y-value for $x = 3$ :
$y = 15 + 9 = 24$

STEP 12

Now plug $x = -2$ into $y = 5x + 9$ :
$y = 5(-2) + 9$

STEP 13

Calculate the y-value for $x = -2$ :
$y = -10 + 9 = -1$

STEP 14

The points of intersection for the first pair of equations are:
$(3, 24) \quad \text{and} \quad (-2, -1)$

STEP 15

Now, let's move on to the second pair of equations:
$y = -x^{2} - 4x + 6$ $y = x - 8$
Again, we will set the two equations equal to each other to find the points of intersection.

STEP 16

Set the two expressions for y equal to each other:
$-x^{2} - 4x + 6 = x - 8$

STEP 17

Add $x^{2}$ and $4x$ to both sides and add 8 to both sides to bring all terms to one side of the equation and set it equal to zero:
$x^{2} + 5x - 14 = 0$

STEP 18

Factor the quadratic equation:
$(x + 7)(x - 2) = 0$

STEP 19

Set each factor equal to zero and solve for x:
$x + 7 = 0 \quad \text{or} \quad x - 2 = 0$

STEP 20

Solve each equation for x:
$x = -7 \quad \text{or} \quad x = 2$

STEP 21

Now that we have the x-values of the points of intersection, we need to find the corresponding y-values. We can do this by plugging the x-values into either of the original equations. We will use $y = x - 8$ .

STEP 22

Plug $x = -7$ into $y = x - 8$ :
$y = (-7) - 8$

STEP 23

Calculate the y-value for $x = -7$ :
$y = -7 - 8 = -15$

STEP 24

Now plug $x = 2$ into $y = x - 8$ :
$y = 2 - 8$

STEP 25

Calculate the y-value for $x = 2$ :
$y = 2 - 8 = -6$

STEP 26

The points of intersection for the second pair of equations are:
$(-7, -15) \quad \text{and} \quad (2, -6)$
The solutions for the given problems are:
a) The points of intersection for the equations $y = x^{2} + 4x + 3$ and $y = 5x + 9$ are $(3, 24)$ and $(-2, -1)$ .
b) The points of intersection for the equations $y = -x^{2} - 4x + 6$ and $y = x - 8$ are $(-7, -15)$ and $(2, -6)$ .

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Find the points of intersection of the given pairs of linear and quadratic equations: y=x2+4x+3,y=5x+9y=x^{2}+4 x+3, y=5 x+9y=x2+4x+3,y=5x+9 and y=−x2−4x+6,y=x−8y=-x^{2}-4 x+6, y=x-8y=−x2−4x+6,y=x−8.

STEP 1

STEP 2

For the first pair of equations, we have:y=x2+4x+3y = x^{2} + 4x + 3y=x2+4x+3 y=5x+9y = 5x + 9y=5x+9To find the points of intersection, we will set the two equations equal to each other, since at the point of intersection, y will have the same value in both equations.

STEP 3

Set the two expressions for y equal to each other:x2+4x+3=5x+9x^{2} + 4x + 3 = 5x + 9x2+4x+3=5x+9

STEP 4

Subtract 5x+95x + 95x+9 from both sides to bring all terms to one side of the equation and set it equal to zero:x2+4x+3−5x−9=0x^{2} + 4x + 3 - 5x - 9 = 0x2+4x+3−5x−9=0

STEP 5

Combine like terms:x2−x−6=0x^{2} - x - 6 = 0x2−x−6=0

STEP 6

Factor the quadratic equation:(x−3)(x+2)=0(x - 3)(x + 2) = 0(x−3)(x+2)=0

STEP 7

Set each factor equal to zero and solve for x:x−3=0orx+2=0x - 3 = 0 \quad \text{or} \quad x + 2 = 0x−3=0orx+2=0

STEP 8

Solve each equation for x:x=3orx=−2x = 3 \quad \text{or} \quad x = -2x=3orx=−2

STEP 9

Now that we have the x-values of the points of intersection, we need to find the corresponding y-values. We can do this by plugging the x-values into either of the original equations. We will use y=5x+9y = 5x + 9y=5x+9.

STEP 10

Plug x=3x = 3x=3 into y=5x+9y = 5x + 9y=5x+9:y=5(3)+9y = 5(3) + 9y=5(3)+9

STEP 11

Calculate the y-value for x=3x = 3x=3:y=15+9=24y = 15 + 9 = 24y=15+9=24

STEP 12

Now plug x=−2x = -2x=−2 into y=5x+9y = 5x + 9y=5x+9:y=5(−2)+9y = 5(-2) + 9y=5(−2)+9

STEP 13

Calculate the y-value for x=−2x = -2x=−2:y=−10+9=−1y = -10 + 9 = -1y=−10+9=−1

STEP 14

The points of intersection for the first pair of equations are:(3,24)and(−2,−1)(3, 24) \quad \text{and} \quad (-2, -1)(3,24)and(−2,−1)

STEP 15

Now, let's move on to the second pair of equations:y=−x2−4x+6y = -x^{2} - 4x + 6y=−x2−4x+6 y=x−8y = x - 8y=x−8Again, we will set the two equations equal to each other to find the points of intersection.

STEP 16

Set the two expressions for y equal to each other:−x2−4x+6=x−8-x^{2} - 4x + 6 = x - 8−x2−4x+6=x−8

STEP 17

Add x2x^{2}x2 and 4x4x4x to both sides and add 8 to both sides to bring all terms to one side of the equation and set it equal to zero:x2+5x−14=0x^{2} + 5x - 14 = 0x2+5x−14=0

STEP 18

Factor the quadratic equation:(x+7)(x−2)=0(x + 7)(x - 2) = 0(x+7)(x−2)=0

STEP 19

Set each factor equal to zero and solve for x:x+7=0orx−2=0x + 7 = 0 \quad \text{or} \quad x - 2 = 0x+7=0orx−2=0

STEP 20

Solve each equation for x:x=−7orx=2x = -7 \quad \text{or} \quad x = 2x=−7orx=2

STEP 21

Now that we have the x-values of the points of intersection, we need to find the corresponding y-values. We can do this by plugging the x-values into either of the original equations. We will use y=x−8y = x - 8y=x−8.

STEP 22

Plug x=−7x = -7x=−7 into y=x−8y = x - 8y=x−8:y=(−7)−8y = (-7) - 8y=(−7)−8

STEP 23

Calculate the y-value for x=−7x = -7x=−7:y=−7−8=−15y = -7 - 8 = -15y=−7−8=−15

STEP 24

Now plug x=2x = 2x=2 into y=x−8y = x - 8y=x−8:y=2−8y = 2 - 8y=2−8

STEP 25

Calculate the y-value for x=2x = 2x=2:y=2−8=−6y = 2 - 8 = -6y=2−8=−6

STEP 26

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Find the points of intersection of the given pairs of linear and quadratic equations: $y=x^{2}+4 x+3, y=5 x+9$ and $y=-x^{2}-4 x+6, y=x-8$ .

For the first pair of equations, we have:
$y = x^{2} + 4x + 3$ $y = 5x + 9$
To find the points of intersection, we will set the two equations equal to each other, since at the point of intersection, y will have the same value in both equations.

Set the two expressions for y equal to each other:
$x^{2} + 4x + 3 = 5x + 9$

Subtract $5x + 9$ from both sides to bring all terms to one side of the equation and set it equal to zero:
$x^{2} + 4x + 3 - 5x - 9 = 0$

Combine like terms:
$x^{2} - x - 6 = 0$

Factor the quadratic equation:
$(x - 3)(x + 2) = 0$

Set each factor equal to zero and solve for x:
$x - 3 = 0 \quad \text{or} \quad x + 2 = 0$

Solve each equation for x:
$x = 3 \quad \text{or} \quad x = -2$

Now that we have the x-values of the points of intersection, we need to find the corresponding y-values. We can do this by plugging the x-values into either of the original equations. We will use $y = 5x + 9$ .

Plug $x = 3$ into $y = 5x + 9$ :
$y = 5(3) + 9$

Calculate the y-value for $x = 3$ :
$y = 15 + 9 = 24$

Now plug $x = -2$ into $y = 5x + 9$ :
$y = 5(-2) + 9$

Calculate the y-value for $x = -2$ :
$y = -10 + 9 = -1$

The points of intersection for the first pair of equations are:
$(3, 24) \quad \text{and} \quad (-2, -1)$

Now, let's move on to the second pair of equations:
$y = -x^{2} - 4x + 6$ $y = x - 8$
Again, we will set the two equations equal to each other to find the points of intersection.

Set the two expressions for y equal to each other:
$-x^{2} - 4x + 6 = x - 8$

Add $x^{2}$ and $4x$ to both sides and add 8 to both sides to bring all terms to one side of the equation and set it equal to zero:
$x^{2} + 5x - 14 = 0$

Factor the quadratic equation:
$(x + 7)(x - 2) = 0$

Set each factor equal to zero and solve for x:
$x + 7 = 0 \quad \text{or} \quad x - 2 = 0$

Solve each equation for x:
$x = -7 \quad \text{or} \quad x = 2$

Now that we have the x-values of the points of intersection, we need to find the corresponding y-values. We can do this by plugging the x-values into either of the original equations. We will use $y = x - 8$ .

Plug $x = -7$ into $y = x - 8$ :
$y = (-7) - 8$

Calculate the y-value for $x = -7$ :
$y = -7 - 8 = -15$

Now plug $x = 2$ into $y = x - 8$ :
$y = 2 - 8$

Calculate the y-value for $x = 2$ :
$y = 2 - 8 = -6$