Respuesta :
Answer:
The equations have one solution at (5, -5).
Step-by-step explanation:
We are given a system of equations:
[tex]\displaystyle{\left \{ {{-2x+5y=-35} \atop {7x+2y=25}} \right.}[/tex]
This system of equations can be solved in three different ways:
- Graphing the equations (method used)
- Substituting values into the equations
- Eliminating variables from the equations
Graphing the Equations
We need to solve each equation and place it in slope-intercept form first. Slope-intercept form is [tex]\text{y = mx + b}[/tex].
Equation 1 is [tex]-2x+5y = -35[/tex]. We need to isolate y.
[tex]\displaystyle{-2x + 5y = -35}\\\\5y = 2x - 35\\\\\frac{5y}{5} = \frac{2x - 35}{5}\\\\y = \frac{2}{5}x - 7[/tex]
Equation 1 is now [tex]y=\frac{2}{5}x-7[/tex].
Equation 2 also needs y to be isolated.
[tex]\displaystyle{7x+2y=25}\\\\2y=-7x+25\\\\\frac{2y}{2}=\frac{-7x+25}{2}\\\\y = -\frac{7}{2}x + \frac{25}{2}[/tex]
Equation 2 is now [tex]y=-\frac{7}{2}x+\frac{25}{2}[/tex].
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.
[tex]\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}[/tex]
[tex]\bullet \ \text{For x = 0,}[/tex]
[tex]\displaystyle{y = \frac{2}{5}(0) - 7}\\\\y = 0 - 7\\\\y = -7[/tex]
[tex]\bullet \ \text{For x = 1,}[/tex]
[tex]\displaystyle{y=\frac{2}{5}(1)-7}\\\\y=\frac{2}{5}-7\\\\y = -\frac{33}{5}[/tex]
[tex]\bullet \ \text{For x = 2,}[/tex]
[tex]\displaystyle{y=\frac{2}{5}(2)-7}\\\\y = \frac{4}{5}-7\\\\y = -\frac{31}{5}[/tex]
[tex]\bullet \ \text{For x = 3,}[/tex]
[tex]\displaystyle{y=\frac{2}{5}(3)-7}\\\\y= \frac{6}{5}-7\\\\y=-\frac{29}{5}[/tex]
[tex]\bullet \ \text{For x = 4,}[/tex]
[tex]\displaystyle{y=\frac{2}{5}(4)-7}\\\\y = \frac{8}{5}-7\\\\y=-\frac{27}{5}[/tex]
[tex]\bullet \ \text{For x = 5,}[/tex]
[tex]\displaystyle{y=\frac{2}{5}(5)-7}\\\\y=2-7\\\\y=-5[/tex]
Now, we can place these values in our table.
[tex]\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}[/tex]
As we can see in our table, the rate of decrease is [tex]-\frac{2}{5}[/tex]. In case we need to determine more values, we can easily either replace x with a new value in the equation or just subtract [tex]-\frac{2}{5}[/tex] from the previous value.
For Equation 2, we need to use the same process. Equation 2 has been resolved to be [tex]y=-\frac{7}{2}x+\frac{25}{2}[/tex]. Therefore, we just use the same process as before to solve for the values.
[tex]\bullet \ \text{For x = 0,}[/tex]
[tex]\displaystyle{y=-\frac{7}{2}(0)+\frac{25}{2}}\\\\y = 0 + \frac{25}{2}\\\\y = \frac{25}{2}[/tex]
[tex]\bullet \ \text{For x = 1,}[/tex]
[tex]\displaystyle{y=-\frac{7}{2}(1)+\frac{25}{2}}\\\\y = -\frac{7}{2} + \frac{25}{2}\\\\y = 9[/tex]
[tex]\bullet \ \text{For x = 2,}[/tex]
[tex]\displaystyle{y=-\frac{7}{2}(2)+\frac{25}{2}}\\\\y = -7+\frac{25}{2}\\\\y = \frac{11}{2}[/tex]
[tex]\bullet \ \text{For x = 3,}[/tex]
[tex]\displaystyle{y=-\frac{7}{2}(3)+\frac{25}{2}}\\\\y = -\frac{21}{2}+\frac{25}{2}\\\\y = 2[/tex]
[tex]\bullet \ \text{For x = 4,}[/tex]
[tex]\displaystyle{y=-\frac{7}{2}(4)+\frac{25}{2}}\\\\y=-14+\frac{25}{2}\\\\y = -\frac{3}{2}[/tex]
[tex]\bullet \ \text{For x = 5,}[/tex]
[tex]\displaystyle{y=-\frac{7}{2}(5)+\frac{25}{2}}\\\\y = -\frac{35}{2}+\frac{25}{2}\\\\y = -5[/tex]
And now, we place these values into the table.
[tex]\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}[/tex]
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
[tex]\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}[/tex] [tex]\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}[/tex]
Therefore, using this data, we have one solution at (5, -5).
