worksheet 7.1
solve each system by graphing.
1) ( x = 4 )
( y = -\frac{1}{2}x + 3 )
2) ( y = x - 2 )
( y = 6x + 3 )
3) ( 2x - y = -1 )
( 2x - y = -4 )
4) ( x - 2y = 6 )
( 5x - 2y = -2 )

Question

Accepted Answer

### Problem 1:
# Explanation:
## Step 1: Analyze the first equation $ x = 4 $
This is a vertical line where all points have an $ x $-coordinate of 4.
## Step 2: Analyze the second equation $ y = -\frac{1}{2}x + 3 $
This is a linear equation in slope - intercept form $ y=mx + b $, where the slope $ m=-\frac{1}{2} $ and the $ y $-intercept $ b = 3 $. To find the intersection with $ x = 4 $, substitute $ x = 4 $ into the second equation.
Substitute $ x = 4 $ into $ y=-\frac{1}{2}x + 3 $:
$ y=-\frac{1}{2}(4)+3=-2 + 3=1 $
# Answer:
The solution of the system $ \begin{cases}x = 4\y=-\frac{1}{2}x + 3\end{cases} $ is $ (4,1) $

### Problem 2:
# Explanation:
## Step 1: Analyze the two linear equations
The first equation $ y=x - 2 $ has a slope $ m = 1 $ and $ y $-intercept $ b=-2 $. The second equation $ y = 6x+3 $ has a slope $ m = 6 $ and $ y $-intercept $ b = 3 $.
To find the intersection, we can set the two equations equal to each other since at the intersection point, the $ y $-values are equal.
Set $ x - 2=6x + 3 $
## Step 2: Solve for $ x $
Subtract $ x $ from both sides: $ - 2=5x+3 $
Subtract 3 from both sides: $ -2-3 = 5x$, so $ -5 = 5x $
Divide both sides by 5: $ x=-1 $
## Step 3: Solve for $ y $
Substitute $ x=-1 $ into $ y=x - 2 $ (we could also use the other equation). Then $ y=-1-2=-3 $
# Answer:
The solution of the system $ \begin{cases}y=x - 2\y = 6x+3\end{cases} $ is $ (-1,-3) $

### Problem 3:
# Explanation:
## Step 1: Rewrite the equations in slope - intercept form
For the first equation $ 2x-y=-1 $, we can rewrite it as $ y=2x + 1 $ (by solving for $ y $: $ y=2x + 1 $)
For the second equation $ 2x-y=-4 $, we can rewrite it as $ y=2x+4 $ (by solving for $ y $: $ y = 2x + 4 $)
## Step 2: Analyze the slopes and $ y $-intercepts
Both lines have the same slope $ m = 2 $ but different $ y $-intercepts ($ b_1 = 1 $ and $ b_2=4 $). Parallel lines (lines with the same slope and different $ y $-intercepts) never intersect.
# Answer:
The system $ \begin{cases}2x - y=-1\2x - y=-4\end{cases} $ has no solution (the lines are parallel)

### Problem 4:
# Explanation:
## Step 1: Rewrite the equations in slope - intercept form
For the first equation $ x-2y=6 $, solve for $ y $:
$ - 2y=-x + 6$, so $ y=\frac{1}{2}x-3 $
For the second equation $ 5x-2y=-2 $, solve for $ y $:
$ -2y=-5x - 2$, so $ y=\frac{5}{2}x + 1 $
## Step 2: Find the intersection point
Set $ \frac{1}{2}x-3=\frac{5}{2}x + 1 $
Subtract $ \frac{1}{2}x $ from both sides: $ -3 = 2x+1 $
Subtract 1 from both sides: $ -4=2x $
Divide both sides by 2: $ x=-2 $
## Step 3: Solve for $ y $
Substitute $ x = - 2 $ into $ y=\frac{1}{2}x-3 $
$ y=\frac{1}{2}(-2)-3=-1-3=-4 $
# Answer:
The solution of the system $ \begin{cases}x - 2y=6\5x - 2y=-2\end{cases} $ is $ (-2,-4) $

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QUESTION IMAGE

Question

Explanation:

Problem 1:

Step 1: Analyze the first equation \( x = 4 \)

Step 2: Analyze the second equation \( y = -\frac{1}{2}x + 3 \)

Step 1: Analyze the two linear equations

Step 2: Solve for \( x \)

Step 3: Solve for \( y \)

Step 1: Rewrite the equations in slope - intercept form

Step 2: Analyze the slopes and \( y \)-intercepts

Answer:

Problem 2: