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:

Write a system of equations to describe the situation below, solve using elimination, and fill in the blanks.\newlineAn employee at a company that assembles chandeliers is packing boxes for shipping. In the first box, he packed 11 small chandelier and 44 large chandeliers, which weighed a total of 207207 pounds. In the second box, he packed 44 small chandeliers and 11 large chandelier, which had a weight of 9393 pounds. Assuming the weight of the box isn't included in the shipping weight, how much does each size of chandelier weigh?\newlineEach small chandelier weighs _\_ pounds and each large one weighs _\_ pounds.

View step-by-step helpView step-by-step help
Homeright chevron icon
Math Problemsright chevron icon
Algebra 2right chevron icon
Solve a system of equations using elimination: word problemsright chevron icon

Full solution

Q. Write a system of equations to describe the situation below, solve using elimination, and fill in the blanks.\newlineAn employee at a company that assembles chandeliers is packing boxes for shipping. In the first box, he packed 11 small chandelier and 44 large chandeliers, which weighed a total of 207207 pounds. In the second box, he packed 44 small chandeliers and 11 large chandelier, which had a weight of 9393 pounds. Assuming the weight of the box isn't included in the shipping weight, how much does each size of chandelier weigh?\newlineEach small chandelier weighs _\_ pounds and each large one weighs _\_ pounds.
  1. Equation 11: Let's denote the weight of each small chandelier as ss and the weight of each large chandelier as ll. The first box with 11 small chandelier and 44 large chandeliers weighs a total of 207207 pounds, which gives us the equation s+4l=207s + 4l = 207.
  2. Equation 22: The second box with 44 small chandeliers and 11 large chandelier weighs 9393 pounds, which gives us the equation 4s+l=934s + l = 93.
  3. Elimination Choice: We now have a system of two equations. To use elimination, we need to eliminate one of the variables, ss or ll. We choose to eliminate ll because its coefficients are multiples of each other.
  4. New Second Equation: To eliminate ll, we multiply the second equation by 44, the coefficient of ll in the first equation. This gives us the new equation 16s+4l=37216s + 4l = 372.
  5. Elimination Step: We now subtract the first equation from the new second equation to eliminate ll. This gives us 16s+4l−(s+4l)=372−20716s + 4l - (s + 4l) = 372 - 207, which simplifies to 15s=16515s = 165.
  6. Calculate Small Chandelier Weight: Dividing both sides of the equation by 1515 gives us s=16515s = \frac{165}{15}, which simplifies to s=11s = 11. So each small chandelier weighs 1111 pounds.
  7. Substitute and Solve: We substitute s=11s = 11 into the first equation and solve for ll. This gives us 11+4l=20711 + 4l = 207. Subtracting 1111 from both sides gives us 4l=1964l = 196.
  8. Calculate Large Chandelier Weight: Dividing both sides of the equation by 44 gives us l=196/4l = 196 / 4, which simplifies to l=49l = 49. So each large chandelier weighs 4949 pounds.

More problems from Solve a system of equations using elimination: word problems

Question
Solve using substitution. 5x−2y=−7 5x - 2y = -7 x=−5 x = -5 (_,_)
Get tutor helpright-arrow

Posted 5 months ago

Question
Is (1,1)(1,1) a solution to this system of equations? \newline4x+10y=144x + 10y = 14\newlinex+6y=7x + 6y = 7\newlineChoices:\newline(A) yes \newline(B) no
Get tutor helpright-arrow

Posted 5 months ago

Question
Which describes the system of equations below?\newliney=–3x+9y = –3x + 9 \newliney=–3x+9y = –3x + 9 \newlineChoices:\newline(A) consistent and independent\text{(A) consistent and independent} \newline(B) consistent and dependent\text{(B) consistent and dependent} \newline(C) inconsistent\text{(C) inconsistent}
Get tutor helpright-arrow

Posted 5 months ago

Question
Solve using elimination. \newline7x−8y=−17 7x - 8y = -17 \newline −7x+3y=2 -7x + 3y = 2 \newline (____,____)(\_\_\_\_, \_\_\_\_)
Get tutor helpright-arrow

Posted 5 months ago

Question
Solve. \newlinex=−2 x = -2 \newline−2x+2y=−8 -2x + 2y = -8 \newline(____,____)(\_\_\_\_, \_\_\_\_)
Get tutor helpright-arrow

Posted 9 months ago

Question
Write a system of equations to describe the situation below, solve using any method, and fill in the blanks.\newlineAt a community barbecue, Mrs. Wilkerson and Mr. Hogan are buying dinner for their families. Mrs. Wilkerson purchases 33 hot dog meals and 33 hamburger meals, paying a total of $36\$36. Mr. Hogan buys 11 hot dog meal and 33 hamburger meals, spending $26\$26 in all. How much do the meals cost?\newlineHot dog meals cost $\$_______ each, and hamburger meals cost $\$________ each.
Get tutor helpright-arrow

Posted 9 months ago

Question
Solve the system of equations by substitution.\newline−3x−y−3z=−11-3x - y - 3z = -11\newlinez=5z = 5\newlinex−y+3z=19x - y + 3z = 19\newline(____.____,____)
Get tutor helpright-arrow

Posted 5 months ago

Question
Solve the system of equations by elimination.\newlinex−3y−2z=10 x - 3y - 2z = 10 \newline3x+2y+2z=14 3x + 2y + 2z = 14 \newline2x−3y−2z=16 2x - 3y - 2z = 16 \newline(_,_,_)(\_,\_,\_)
Get tutor helpright-arrow

Posted 5 months ago

Question
Solve the system of equations.\newliney=x2+36x+3 y = x^2 + 36x + 3 \newliney=22x−37 y = 22x - 37 \newlineWrite the coordinates in exact form. Simplify all fractions and radicals.\newline(_,_)(\_,\_)\newline(_,_)(\_,\_)
Get tutor helpright-arrow

Posted 5 months ago

Question
Solve the system of equations.\newliney=−x−24 y = -x - 24 \newlinex2+y2=488 x^2 + y^2 = 488 \newlineWrite the coordinates in exact form. Simplify all fractions and radicals.\newline(_,_) (\_,\_) \newline(_,_) (\_,\_)
Get tutor helpright-arrow

Posted 5 months ago

View all (43)

Related topics

Algebra - Order of OperationsAlgebra - Distributive Property`X` and `Y` AxesGeometry - Scalene TriangleCommon MultipleGeometry - Quadrant