Question

Self CheckFor the following problems, the answers will be given. Show the calculations that needed to get to the answerand explain each step.ProblemTwo small pitchers and one large pitcher can hold 8 cups of water.One large pitcher minus one small pitcher constitutes 2 cups ofwater.How many cups of water can each pitcher hold?Answer• small pitcher = 2 cups ofwaterlarge pitcher = 4 cups ofwater.Write equations to represent this situation.Use x to represent the cups of water that the small pitcher can hold and y to represent the cups thathe large pitcher can hold.Equation 1: Two small pitchers and one large pitcher can hold 8 cups of water.

Answer 1

SOLUTION

From the question x represents what small pitcher can hold and y what larger pitcher can hold

Two small pitchers and one large pitcher can hold 8 cups of water.

This is written as

$\begin{gathered} x+x+y=8\text{ cups of water } \\ x+x+y=8 \\ 2x+y=8\ldots\ldots\ldots\text{.equation 1} \end{gathered}$

One large pitcher minus one small pitcher constitutes 2 cups of water.

This is written as

$\begin{gathered} y-x=2\text{ cups of water } \\ y-x=2\ldots\ldots\ldots\text{.equation 2} \end{gathered}$

Solving the systems of simultaneous equation, we have

$\begin{gathered} 2x+y=8 \\ y-x=2 \\ \text{arranging we have } \\ 2x+y=8 \\ -x+y=2 \\ m\text{ultiplying the second equation by minus sign we have } \\ 2x+y=8 \\ -(-x+y=2 \\ 2x+y=8 \\ x-y=-2 \end{gathered}$

Eliminating y, we have

$\begin{gathered} (2x+x)+(y-y)=8-2 \\ 3x+0=6 \\ 3x=6 \\ x=\frac{6}{3} \\ x=2 \end{gathered}$

Substituting the x for 2 into equation 1, we have

$\begin{gathered} 2x+y=8 \\ 2(2)+y=8 \\ 4+y=8 \\ y=8-4 \\ y=4 \end{gathered}$

So, the answer is x = 2 and y = 4.

Hence small pitcher = 2 cups of water and large pitcher = 4 cups of water