Write a system of equations to describe the situation below, solve using elimination, and fill in the blanks. $\newline$ Friends and family of the bride will be helping assemble centerpieces for the wedding reception. On the right side of the room, there will be $6$ round tables and $2$ rectangular tables, which will require a total of $14$ centerpieces. On the left side, there will be $6$ round tables and $1$ rectangular table, for which they will need to assemble a total of $10$ centerpieces. How many centerpieces will be on each table? $\newline$ There will be $\_$ centerpieces on every round table and $\_$ centerpieces on every rectangular one.

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Algebra 1

Solve a system of equations using elimination: word problems

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Q. Write a system of equations to describe the situation below, solve using elimination, and fill in the blanks.

\newline

Friends and family of the bride will be helping assemble centerpieces for the wedding reception. On the right side of the room, there will be

6

round tables and

2

rectangular tables, which will require a total of

14

centerpieces. On the left side, there will be

6

round tables and

1

rectangular table, for which they will need to assemble a total of

10

centerpieces. How many centerpieces will be on each table?

\newline

There will be

\_

centerpieces on every round table and

\_

centerpieces on every rectangular one.

Define variables: Let's define variables for the number of centerpieces on each type of table. $\newline$ Let $x$ be the number of centerpieces on each round table. $\newline$ Let $y$ be the number of centerpieces on each rectangular table. $\newline$ We can now write two equations based on the given information.
Write equations for right side: Write the equation for the right side of the room. $\newline$ $6$ round tables and $2$ rectangular tables require $14$ centerpieces. $\newline$ This gives us the equation: $6x + 2y = 14$ .
Write equations for left side: Write the equation for the left side of the room. $\newline$ $6$ round tables and $1$ rectangular table require $10$ centerpieces. $\newline$ This gives us the equation: $6x + y = 10$ .
Eliminate variable $y$ : We have the system of equations: $\newline$ $6x + 2y = 14$ $\newline$ $6x + y = 10$ $\newline$ We need to eliminate one of the variables. We can eliminate $y$ by multiplying the second equation by $-2$ and then adding it to the first equation.
Multiply second equation by $-2$ : Multiply the second equation by $-2＂): \$-2(6x + y) = -2(10)$ $-12x - 2y = -20$ Now we have the new system of equations: $6x + 2y = 14$ $-12x - 2y = -20$
Add equations to eliminate y: Add the two equations to eliminate y: $\newline$ $(6x + 2y) + (-12x - 2y) = 14 + (-20)$ $\newline$ $6x - 12x + 2y - 2y = 14 - 20$ $\newline$ $-6x = -6$
Solve for x: Solve for x: $\newline$ $-6x = -6$ $\newline$ Divide both sides by $-6$ : $\newline$ $x = 1$ $\newline$ So, there is $1$ centerpiece on each round table.
Substitute $x$ into second equation: Substitute $x = 1$ into the second original equation to solve for $y$ : $6x + y = 10$ $6(1) + y = 10$ $6 + y = 10$ $y = 10 - 6$ $y = 4$ So, there are $4$ centerpieces on each rectangular table.