Solve the following system of equations graphically on the set of axes below.

y, equals, one fifth, x, plus, 6
y=
5
1

x+6
4, x, plus, 5, y, equals, 5
4x+5y=5

3 answers

To solve the system of equations graphically, we will represent both equations on a coordinate plane.

  1. The first equation is: \[ y = \frac{1}{5}x + 6 \] This is a linear equation in slope-intercept form (y = mx + b) where the slope (m) is \(\frac{1}{5}\) and the y-intercept (b) is 6.

    To graph this line:

    • Start at the y-intercept (0, 6).
    • From the point (0, 6), use the slope \(\frac{1}{5}\) to find another point. For example, for every 5 units you move to the right (increase in x), you move up 1 unit (increase in y), leading to the point (5, 7).
    • Plot these points and draw a line through them.
  2. The second equation is: \[ 4x + 5y = 5 \] To graph this equation, we can solve for y: \[ 5y = 5 - 4x \implies y = -\frac{4}{5}x + 1 \] This is also in slope-intercept form, where the slope is \(-\frac{4}{5}\) and the y-intercept is 1.

    To graph this line:

    • Start at the y-intercept (0, 1).
    • From (0, 1), utilize the slope \(-\frac{4}{5}\). For every 5 units you move to the right, move down 4 units, which gives you the point (5, -3).
    • Plot these points and draw a line through them.

Finding the intersection:

To find the solution to the system of equations, we need to determine where the two lines intersect. You can do this by visually analyzing the graph or algebraically by setting the two equations equal to each other.

Solving the equations:

Set the two expressions for y equal to each other: \[ \frac{1}{5}x + 6 = -\frac{4}{5}x + 1 \]

To eliminate fractions, multiply through by 5: \[ x + 30 = -4x + 5 \]

Combine like terms: \[ x + 4x = 5 - 30 \] \[ 5x = -25 \] \[ x = -5 \]

Now, substitute \(x = -5\) back into one of the original equations to find y. Using the first equation: \[ y = \frac{1}{5}(-5) + 6 = -1 + 6 = 5 \]

Conclusion:

The solution to the system of equations is: \[ (x, y) = (-5, 5) \]

So, the point of intersection can be plotted on your graph to show the solution of the system of equations.

4x+5y=5

Let's go through the problem again with just the system of equations you've provided, focusing on both equations clearly. The system includes:

  1. Equation 1: \[ y = \frac{1}{5}x + 6 \]

  2. Equation 2: \[ 4x + 5y = 5 \]

Step 1: Graph the First Equation

For the equation \(y = \frac{1}{5}x + 6\):

  • Y-intercept (0,6): Start at the point \((0, 6)\) on the y-axis.

  • Using the slope: The slope \(\frac{1}{5}\) means that for every 5 units you move to the right (increase in \(x\)), you move up 1 unit (increase in \(y\)). So from (0, 6) you can find another point:

    • Move right 5 to \(x = 5\), and up 1 to find the point (5, 7).

Step 2: Graph the Second Equation

For the equation \(4x + 5y = 5\):

First, we need to rearrange this equation into slope-intercept form (\(y = mx + b\)): \[ 5y = 5 - 4x \] \[ y = -\frac{4}{5}x + 1 \]

Now, graph \(y = -\frac{4}{5}x + 1\):

  • Y-intercept (0,1): Start at \((0, 1)\) on the y-axis.

  • Using the slope: The slope \(-\frac{4}{5}\) means for every 5 units you move to the right, you move down 4 units. So from (0, 1):

    • Move right 5 to \(x = 5\) and down 4 to find the point (5, -3).

Step 3: Find the Intersection

To find the intersection of these two lines, we can either graph them and see where they intersect, or solve the equations algebraically.

Set the two equations equal to each other:

\[ \frac{1}{5}x + 6 = -\frac{4}{5}x + 1 \]

Clear fractions by multiplying the entire equation by 5:

\[ x + 30 = -4x + 5 \]

Rearrange the equation:

\[ x + 4x = 5 - 30 \] \[ 5x = -25 \] \[ x = -5 \]

Find \(y\) by substituting \(x = -5\) back into one of the original equations. Let's use \(y = \frac{1}{5}x + 6\):

\[ y = \frac{1}{5}(-5) + 6 = -1 + 6 = 5 \]

Conclusion

The solution to the system of equations is: \[ (x, y) = (-5, 5) \]

On a graph, you would plot the two lines and find that they intersect at the point (-5, 5). This point is the solution to the system of equations.