Resourcesdown chevron
Testimonials
Plans
Sign in
Sign up
Resourcesright arrow
Testimonials
Plans
Bytelearn - cat image with glassesAI tutor

Welcome to Bytelearn!

Let’s check out your problem:

Solve the system of equations.\newliney=x2+28x−41y = x^2 + 28x - 41\newliney=28x−16y = 28x - 16\newlineWrite the coordinates in exact form. Simplify all fractions and radicals.\newline(______,______)\newline(______,______)

View step-by-step helpView step-by-step help
Homeright chevron icon
Math Problemsright chevron icon
Algebra 1right chevron icon
Solve a system of linear and quadratic equationsright chevron icon

Full solution

Q. Solve the system of equations.\newliney=x2+28x−41y = x^2 + 28x - 41\newliney=28x−16y = 28x - 16\newlineWrite the coordinates in exact form. Simplify all fractions and radicals.\newline(______,______)\newline(______,______)
  1. Set Equations Equal: We have the system of equations:\newliney=x2+28x−41y = x^2 + 28x - 41\newliney=28x−16y = 28x - 16\newlineTo find the intersection points, set the two equations equal to each other.\newlinex2+28x−41=28x−16x^2 + 28x - 41 = 28x - 16
  2. Simplify Equation: Simplify the equation by subtracting 28x28x from both sides and adding 1616 to both sides.\newlinex2+28x−41−28x+16=28x−16−28x+16x^2 + 28x - 41 - 28x + 16 = 28x - 16 - 28x + 16\newlinex2−25=0x^2 - 25 = 0
  3. Solve Quadratic Equation: Solve the simplified quadratic equation for xx.x2−25=(x−5)(x+5)x^2 - 25 = (x - 5)(x + 5)Set each factor equal to zero and solve for xx.(x−5)=0 or (x+5)=0(x - 5) = 0 \text{ or } (x + 5) = 0x=5 or x=−5x = 5 \text{ or } x = -5
  4. Find Y-Values: Find the corresponding y-values for each x-value by substituting back into one of the original equations. We'll use y=28x−16y = 28x - 16.\newlineFor x=5x = 5:\newliney=28(5)−16y = 28(5) - 16\newliney=140−16y = 140 - 16\newliney=124y = 124
  5. Write Coordinates: For x=−5x = -5: \newliney=28(−5)−16y = 28(-5) - 16 \newliney=−140−16y = -140 - 16 \newliney=−156y = -156
  6. Write Coordinates: For x=−5x = -5:y=28(−5)−16y = 28(-5) - 16y=−140−16y = -140 - 16y=−156y = -156Write the coordinates in exact form. The first coordinate is (5,124)(5, 124). The second coordinate is (−5,−156)(-5, -156).

More problems from Solve a system of linear and quadratic equations

Question
Solve for v v .\newlinev2+8v+12=0 v^2+8v+12=0\newlineWrite each solution as an integer, proper fraction, or improper fraction in simplest form. If there are multiple solutions, separate them with commas.\newlinev= v=_____
Get tutor helpright-arrow

Posted 1 month ago

Question
Solve for u u .\newlineu2=9 u^2 = 9 \newlineWrite your answers as integers or as proper or improper fractions in simplest form.\newlineu= u = _____ or u= u = _____
Get tutor helpright-arrow

Posted 2 months ago

Question
Complete the square. Fill in the number that makes the polynomial a perfect-square quadratic.\newlinek2−20k+_____k^2 - 20k + \_\_\_\_\_
Get tutor helpright-arrow

Posted 2 months ago

Question
Solve by completing the square.\newlineu2−24u−23=0u^2 - 24u - 23 = 0\newlineWrite your answers as integers, proper or improper fractions in simplest form, or decimals rounded to the nearest hundredth.\newline`u` = ________ or `u` =______
Get tutor helpright-arrow

Posted 2 months ago

Question
Find the discriminant. \newline2q2−9q+8=02q^2 - 9q + 8 = 0\newline______
Get tutor helpright-arrow

Posted 2 months ago

Question
Solve the system of equations.\newliney=22x−38 y = 22x - 38 \newliney=x2+11x−14 y = x^2 + 11x - 14 \newlineWrite the coordinates in exact form. Simplify all fractions and radicals.\newline(_,_) (\_,\_) \newline(_,_) (\_,\_)
Get tutor helpright-arrow

Posted 1 month ago

Question
Solve for aa.\newline(a+5)(a+1)=0(a + 5)(a + 1) = 0\newlineWrite your answers as integers or as proper or improper fractions in simplest form.\newlinea=a = _____ or a=a = _____
Get tutor helpright-arrow

Posted 2 months ago

Question
Let p=x2−7 p=x^{2}-7 . Which equation is equivalent to (x2−7)2−4x2+28=5 (x^{2}-7)^{2}-4x^{2}+28=5 in terms of p p ? Choose 11 answer: \newline(A) p2−4p+23=0 p^{2}-4p+23=0 \newline(B) p2+4p−5=0 p^{2}+4p-5=0 \newline(C) p2−4p−5=0 p^{2}-4p-5=0 \newline(D) p2+4p+23=0 p^{2}+4p+23=0
Get tutor helpright-arrow

Posted 2 months ago

Question
Let m=2x+3m=2x+3. Which equation is equivalent to (2x+3)2−14x−21=−6(2x+3)^{2}-14x-21=-6 in terms of mm? Choose 11 answer: \newline(A) m2+7m+6=0m^{2}+7m+6=0 \newline(B) m2−7m−15=0m^{2}-7m-15=0 \newline(C) m2−7m+6=0m^{2}-7m+6=0 \newline(D) m2+7m−15=0m^{2}+7m-15=0
Get tutor helpright-arrow

Posted 2 months ago

Question
Let m=x2+3m=x^{2}+3. Which equation is equivalent to (x2+3)2+7x2+21=−10(x^{2}+3)^{2}+7x^{2}+21=-10 in terms of mm? Choose 11 answer:\newline(A) m2−7m+31=0m^{2}-7m+31=0\newline(B) m2+7m+31=0m^{2}+7m+31=0\newline(C) m2−7m+10=0m^{2}-7m+10=0\newline(D) m2+7m+10=0m^{2}+7m+10=0
Get tutor helpright-arrow

Posted 2 months ago

View all (28)