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

Question

y, equals, one half, x, minus, 5
y=
2
1
​
x−5
x, plus, y, equals, 1
x+y=1

Answer 1

To solve the system of equations graphically, we will plot both equations on the same set of axes.

1. First Equation:

\( y = \frac{1}{2}x - 5 \)

This is in slope-intercept form \( y = mx + b \), where \( m = \frac{1}{2} \) and \( b = -5 \).

• Y-Intercept: The point where the line crosses the y-axis (when \( x = 0 \)): \[ y = \frac{1}{2}(0) - 5 = -5 \] So the y-intercept is (0, -5).

• X-Intercept: The point where the line crosses the x-axis (when \( y = 0 \)): \[ 0 = \frac{1}{2}x - 5 \implies \frac{1}{2}x = 5 \implies x = 10 \] So the x-intercept is (10, 0).

2. Second Equation:

\( x + y = 1 \)

We can rearrange this equation to slope-intercept form \( y = mx + b \): \[ y = -x + 1 \]

• Y-Intercept: When \( x = 0 \): \[ y = -0 + 1 = 1 \] So the y-intercept is (0, 1).

• X-Intercept: When \( y = 0 \): \[ 0 = -x + 1 \implies x = 1 \] So the x-intercept is (1, 0).

Plotting the Lines

Now, let's plot the intercepts for each equation:

1. The line for \( y = \frac{1}{2}x - 5 \) will pass through:

   • (0, -5)
   • (10, 0)

2. The line for \( y = -x + 1 \) will pass through:

   • (0, 1)
   • (1, 0)

Finding the Point of Intersection

The solution to the system of equations is where the two lines intersect.

To find this point algebraically, set the right-hand sides of the equations equal to each other:

\[ \frac{1}{2}x - 5 = -x + 1 \]

Multiplying through by 2 to eliminate the fraction:

\[ x - 10 = -2x + 2 \]

Combine like terms:

\[ 3x = 12 \implies x = 4 \]

Substituting \( x = 4 \) back into one of the original equations to find \( y \):

Using \( y = \frac{1}{2}x - 5 \):

\[ y = \frac{1}{2}(4) - 5 = 2 - 5 = -3 \]

Intersection Point

So, the solution to the system of equations is: \[ (4, -3) \]

You can then plot both equations on the axes, and the point (4, -3) will be where they intersect.

Answer 2

x+y=1

Answer 3

Let's clarify and analyze the given equations again, specifically focusing on the one you just mentioned, which is part of the system.

Given Equations

1. \( y = \frac{1}{2}x - 5 \)
2. \( x + y = 1 \) (or rearranging gives \( y = -x + 1 \))

We will solve this system graphically by finding the lines corresponding to each equation and determining the intersection.

Step 1: Graph the First Equation

For \( y = \frac{1}{2}x - 5 \):

• Y-Intercept: When \( x = 0 \): \[ y = \frac{1}{2}(0) - 5 = -5 \] So the point is (0, -5).

• X-Intercept: When \( y = 0 \): \[ 0 = \frac{1}{2}x - 5 \implies \frac{1}{2}x = 5 \implies x = 10 \] So the point is (10, 0).

Step 2: Graph the Second Equation

For \( x + y = 1 \), or rearranged as \( y = -x + 1 \):

• Y-Intercept: When \( x = 0 \): \[ y = -0 + 1 = 1 \] So the point is (0, 1).

• X-Intercept: When \( y = 0 \): \[ 0 = -x + 1 \implies x = 1 \] So the point is (1, 0).

Step 3: Plotting the Lines

• The first line \( y = \frac{1}{2}x - 5 \) will pass through:

  • (0, -5)
  • (10, 0)

• The second line \( y = -x + 1 \) will pass through:

  • (0, 1)
  • (1, 0)

Step 4: Finding the Intersection Point

To find the intersection algebraically, set the right sides equal: \[ \frac{1}{2}x - 5 = -x + 1 \]

Multiply through by 2 to eliminate the fraction: \[ x - 10 = -2x + 2 \]

Combine like terms: \[ 3x = 12 \implies x = 4 \]

Now substitute \( x = 4 \) back into either equation to find \( y \): Using \( y = -x + 1 \): \[ y = -4 + 1 = -3 \]

The Intersection Point

The solution to the system of equations is: \[ (4, -3) \]

Summary

Graphically, the lines will intersect at the point (4, -3). If you plot both lines on a graph, this intersection point will be where both equations are satisfied simultaneously.