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Pachacha [2.7K]
2 years ago
7

Joanie bought a total of 56 presents to put in 7 gift bags.She put 3 pencils and an equal number of stickers in each bag.How man

y stickers did she put in each bag​
Mathematics
1 answer:
amid [387]2 years ago
7 0

Answer:

5 stickers were put in each bag.

Step-by-step explanation:

There is a total of 56 presents and 7 gift bags. If she puts three pencils in each bag, 7 times 3 is 21. 56 presents minus 21 pencils is 35 presents left. 35 divided by 7 is 5, so there are 5 stickers in each bag.

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Solve the system of equations.<br><br><br><br> −2x+5y =−35<br> 7x+2y =25
Otrada [13]

Answer:

The equations have one solution at (5, -5).

Step-by-step explanation:

We are given a system of equations:

\displaystyle{\left \{ {{-2x+5y=-35} \atop {7x+2y=25}} \right.}

This system of equations can be solved in three different ways:

  1. Graphing the equations (method used)
  2. Substituting values into the equations
  3. Eliminating variables from the equations

<u>Graphing the Equations</u>

We need to solve each equation and place it in slope-intercept form first. Slope-intercept form is \text{y = mx + b}.

Equation 1 is -2x+5y = -35. We need to isolate y.

\displaystyle{-2x + 5y = -35}\\\\5y = 2x - 35\\\\\frac{5y}{5} = \frac{2x - 35}{5}\\\\y = \frac{2}{5}x - 7

Equation 1 is now y=\frac{2}{5}x-7.

Equation 2 also needs y to be isolated.

\displaystyle{7x+2y=25}\\\\2y=-7x+25\\\\\frac{2y}{2}=\frac{-7x+25}{2}\\\\y = -\frac{7}{2}x + \frac{25}{2}

Equation 2 is now y=-\frac{7}{2}x+\frac{25}{2}.

Now, we can graph both of these using a data table and plotting points on the graph. If the two lines intersect at a point, this is a solution for the system of equations.

The table below has unsolved y-values - we need to insert the value of x and solve for y and input these values in the table.

\begin{array}{|c|c|} \cline{1-2} \textbf{x} & \textbf{y} \\ \cline{1-2} 0 & a \\ \cline{1-2} 1 & b \\ \cline{1-2} 2 & c \\ \cline{1-2} 3 & d \\ \cline{1-2} 4 & e \\ \cline{1-2} 5 & f \\ \cline{1-2} \end{array}

\bullet \ \text{For x = 0,}

\displaystyle{y = \frac{2}{5}(0) - 7}\\\\y = 0 - 7\\\\y = -7

\bullet \ \text{For x = 1,}

\displaystyle{y=\frac{2}{5}(1)-7}\\\\y=\frac{2}{5}-7\\\\y = -\frac{33}{5}

\bullet \ \text{For x = 2,}

\displaystyle{y=\frac{2}{5}(2)-7}\\\\y = \frac{4}{5}-7\\\\y = -\frac{31}{5}

\bullet \ \text{For x = 3,}

\displaystyle{y=\frac{2}{5}(3)-7}\\\\y= \frac{6}{5}-7\\\\y=-\frac{29}{5}

\bullet \ \text{For x = 4,}

\displaystyle{y=\frac{2}{5}(4)-7}\\\\y = \frac{8}{5}-7\\\\y=-\frac{27}{5}

\bullet \ \text{For x = 5,}

\displaystyle{y=\frac{2}{5}(5)-7}\\\\y=2-7\\\\y=-5

Now, we can place these values in our table.

\begin{array}{|c|c|} \cline{1-2} \textbf{x} & \textbf{y} \\ \cline{1-2} 0 & -7 \\ \cline{1-2} 1 & -33/5 \\ \cline{1-2} 2 & -31/5 \\ \cline{1-2} 3 & -29/5 \\ \cline{1-2} 4 & -27/5 \\ \cline{1-2} 5 & -5 \\ \cline{1-2} \end{array}

As we can see in our table, the rate of decrease is -\frac{2}{5}. In case we need to determine more values, we can easily either replace x with a new value in the equation or just subtract -\frac{2}{5} from the previous value.

For Equation 2, we need to use the same process. Equation 2 has been resolved to be y=-\frac{7}{2}x+\frac{25}{2}. Therefore, we just use the same process as before to solve for the values.

\bullet \ \text{For x = 0,}

\displaystyle{y=-\frac{7}{2}(0)+\frac{25}{2}}\\\\y = 0 + \frac{25}{2}\\\\y = \frac{25}{2}

\bullet \ \text{For x = 1,}

\displaystyle{y=-\frac{7}{2}(1)+\frac{25}{2}}\\\\y = -\frac{7}{2} + \frac{25}{2}\\\\y = 9

\bullet \ \text{For x = 2,}

\displaystyle{y=-\frac{7}{2}(2)+\frac{25}{2}}\\\\y = -7+\frac{25}{2}\\\\y = \frac{11}{2}

\bullet \ \text{For x = 3,}

\displaystyle{y=-\frac{7}{2}(3)+\frac{25}{2}}\\\\y = -\frac{21}{2}+\frac{25}{2}\\\\y = 2

\bullet \ \text{For x = 4,}

\displaystyle{y=-\frac{7}{2}(4)+\frac{25}{2}}\\\\y=-14+\frac{25}{2}\\\\y = -\frac{3}{2}

\bullet \ \text{For x = 5,}

\displaystyle{y=-\frac{7}{2}(5)+\frac{25}{2}}\\\\y = -\frac{35}{2}+\frac{25}{2}\\\\y = -5

And now, we place these values into the table.

\begin{array}{|c|c|} \cline{1-2} \textbf{x} & \textbf{y} \\ \cline{1-2} 0 & 25/2 \\ \cline{1-2} 1 & 9 \\ \cline{1-2} 2 & 11/2 \\ \cline{1-2} 3 & 2 \\ \cline{1-2} 4 & -3/2 \\ \cline{1-2} 5 & -5 \\ \cline{1-2} \end{array}

When we compare our two tables, we can see that we have one similarity - the points are the same at x = 5.

Equation 1                  Equation 2

\begin{array}{|c|c|} \cline{1-2} \textbf{x} & \textbf{y} \\ \cline{1-2} 0 & -7 \\ \cline{1-2} 1 & -33/5 \\ \cline{1-2} 2 & -31/5 \\ \cline{1-2} 3 & -29/5 \\ \cline{1-2} 4 & -27/5 \\ \cline{1-2} 5 & -5 \\ \cline{1-2} \end{array}                 \begin{array}{|c|c|} \cline{1-2} \textbf{x} & \textbf{y} \\ \cline{1-2} 0 & 25/2 \\ \cline{1-2} 1 & 9 \\ \cline{1-2} 2 & 11/2 \\ \cline{1-2} 3 & 2 \\ \cline{1-2} 4 & -3/2 \\ \cline{1-2} 5 & -5 \\ \cline{1-2} \end{array}

Therefore, using this data, we have one solution at (5, -5).

4 0
3 years ago
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