Resources

Resources

Bytelearn - cat image with glasses

AI tutor

Welcome to Bytelearn!

Let’s check out your problem:

Solve the system of equations. $\newline$ $y = 12x^2 - x - 19$ $\newline$ $y = -x + 29$ $\newline$ Write the coordinates in exact form. Simplify all fractions and radicals. $\newline$ (,) $\newline$ (,)

View step-by-step help

View step-by-step help

Solve a system of linear and quadratic equations

Full solution

Q. Solve the system of equations.

\newline

y = 12x^2 - x - 19

\newline

y = -x + 29

\newline

Write the coordinates in exact form. Simplify all fractions and radicals.

\newline

(______,______)

\newline

(______,______)

Set Equations Equal: We have the following system of equations: $\newline$ $y = 12x^2 - x - 19$ $\newline$ $y = -x + 29$ $\newline$ To find the solution, we will set the two equations equal to each other since they both equal $y$ . $\newline$ $12x^2 - x - 19 = -x + 29$
Simplify and Rearrange: Simplify the equation by moving all terms to one side to set the equation to zero. $\newline$ $12x^2 - x - 19 + x - 29 = 0$ $\newline$ $12x^2 - 48 = 0$
Factor and Solve for $x$ : Solve for $x$ by factoring or using the quadratic formula. In this case, we can factor the equation. $12x^2 - 48 = 0$ Divide by $12$ to simplify: $x^2 - 4 = 0$ Factor the difference of squares: $(x - 2)(x + 2) = 0$
Substitute $x$ into Equation: Solve for $x$ by setting each factor equal to zero. $\newline$ $x - 2 = 0$ or $x + 2 = 0$ $\newline$ $x = 2$ or $x = -2$
Write Coordinates: Find the corresponding $y$ -values by substituting $x$ back into one of the original equations. We can use $y = -x + 29$ .
For $x = 2$ :
$y = -(2) + 29$
$y = 27$
For $x = -2$ :
$y = -(-2) + 29$
$y = 31$
Write Coordinates: Find the corresponding $y$ -values by substituting $x$ back into one of the original equations. We can use $y = -x + 29$ . For $x = 2$ : $y = -(2) + 29$ $y = 27$ For $x = -2$ : $y = -(-2) + 29$ $y = 31$ Write the coordinates in exact form. The solutions to the system of equations are the points where the two graphs intersect, which are the $x$ -values we found and their corresponding $y$ -values. First Coordinate: $x$ $1$ Second Coordinate: $x$ $2$

More problems from Solve a system of linear and quadratic equations

Question

v

\newline

v^2+8v+12=0

\newline

Write each solution as an integer, proper fraction, or improper fraction in simplest form. If there are multiple solutions, separate them with commas.

\newline

v=

Posted 1 month ago

Question

u

\newline

u^2 = 9

\newline

Write your answers as integers or as proper or improper fractions in simplest form.

\newline

u =

u =

Posted 2 months ago

Question

Complete the square. Fill in the number that makes the polynomial a perfect-square quadratic.

\newline

k^2 - 20k + \_\_\_\_\_

Posted 2 months ago

Question

Solve by completing the square.

\newline

u^2 - 24u - 23 = 0

\newline

Write your answers as integers, proper or improper fractions in simplest form, or decimals rounded to the nearest hundredth.

\newline

`u` = ________ or `u` =______

Posted 2 months ago

Question

Find the discriminant.

\newline

2q^2 - 9q + 8 = 0

\newline

Posted 2 months ago

Question

Solve the system of equations.

\newline

y = 22x - 38

\newline

y = x^2 + 11x - 14

\newline

Write the coordinates in exact form. Simplify all fractions and radicals.

\newline

(\_,\_)

\newline

(\_,\_)

Posted 1 month ago

Question

a

\newline

(a + 5)(a + 1) = 0

\newline

Write your answers as integers or as proper or improper fractions in simplest form.

\newline

a =

a =

Posted 2 months ago

Question

p=x^{2}-7

. Which equation is equivalent to

(x^{2}-7)^{2}-4x^{2}+28=5

p

1

\newline

p^{2}-4p+23=0

\newline

p^{2}+4p-5=0

\newline

p^{2}-4p-5=0

\newline

p^{2}+4p+23=0

Posted 2 months ago

Question

m=2x+3

. Which equation is equivalent to

(2x+3)^{2}-14x-21=-6

m

1

\newline

m^{2}+7m+6=0

\newline

m^{2}-7m-15=0

\newline

m^{2}-7m+6=0

\newline

m^{2}+7m-15=0

Posted 2 months ago

Question

m=x^{2}+3

. Which equation is equivalent to

(x^{2}+3)^{2}+7x^{2}+21=-10

m

1

\newline

m^{2}-7m+31=0

\newline

m^{2}+7m+31=0

\newline

m^{2}-7m+10=0

\newline

m^{2}+7m+10=0

Posted 2 months ago