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Sagot :
To find the correct graph of the given system of equations, we need to follow these steps:
1. Rewrite each equation in slope-intercept form (y = mx + b) for easier graphing.
2. Graph each equation on a coordinate plane.
3. Find the intersection point of the two lines, which represents the solution to the system of equations.
Let's solve each equation step-by-step:
### First equation: [tex]\( y - 2x = -1 \)[/tex]
1. Add [tex]\( 2x \)[/tex] to both sides:
[tex]\[ y = 2x - 1 \][/tex]
This is already in slope-intercept form, where the slope (m) is 2 and the y-intercept (b) is -1.
### Second equation: [tex]\( x + 3y = 4 \)[/tex]
1. Subtract [tex]\( x \)[/tex] from both sides:
[tex]\[ 3y = -x + 4 \][/tex]
2. Divide by 3:
[tex]\[ y = -\frac{1}{3}x + \frac{4}{3} \][/tex]
This is in slope-intercept form, where the slope (m) is [tex]\( -\frac{1}{3} \)[/tex] and the y-intercept (b) is [tex]\( \frac{4}{3} \)[/tex].
### Graphing the equations:
1. First equation [tex]\( y = 2x - 1 \)[/tex]:
- The y-intercept is -1, so the line crosses the y-axis at -1.
- The slope is 2, which means for every 1 unit increase in x, y increases by 2 units.
- Points to plot: (0, -1), (1, 1), (2, 3), etc.
2. Second equation [tex]\( y = -\frac{1}{3}x + \frac{4}{3} \)[/tex]:
- The y-intercept is [tex]\( \frac{4}{3} \approx 1.33 \)[/tex].
- The slope is [tex]\( -\frac{1}{3} \)[/tex], which means for every 1 unit increase in x, y decreases by [tex]\( \frac{1}{3} \)[/tex] units.
- Points to plot: (0, [tex]\( \frac{4}{3} \)[/tex]), (3, 1), (-3, 2), etc.
### Intersection Point:
The intersection point of the two lines can be solved algebraically or visually from the graph. After solving, we find that:
[tex]\[ x = 1, \quad y = 1 \][/tex]
Therefore, the correct graph will show two lines intersecting at the point (1, 1). The graph should display:
- A line passing through points such as (0, -1) and (1, 1) for the first equation.
- Another line passing through points such as (0, [tex]\( \frac{4}{3} \)[/tex]) and (3, 1) for the second equation.
Choose the graph that matches these characteristics, with the intersection point clearly at (1, 1).
1. Rewrite each equation in slope-intercept form (y = mx + b) for easier graphing.
2. Graph each equation on a coordinate plane.
3. Find the intersection point of the two lines, which represents the solution to the system of equations.
Let's solve each equation step-by-step:
### First equation: [tex]\( y - 2x = -1 \)[/tex]
1. Add [tex]\( 2x \)[/tex] to both sides:
[tex]\[ y = 2x - 1 \][/tex]
This is already in slope-intercept form, where the slope (m) is 2 and the y-intercept (b) is -1.
### Second equation: [tex]\( x + 3y = 4 \)[/tex]
1. Subtract [tex]\( x \)[/tex] from both sides:
[tex]\[ 3y = -x + 4 \][/tex]
2. Divide by 3:
[tex]\[ y = -\frac{1}{3}x + \frac{4}{3} \][/tex]
This is in slope-intercept form, where the slope (m) is [tex]\( -\frac{1}{3} \)[/tex] and the y-intercept (b) is [tex]\( \frac{4}{3} \)[/tex].
### Graphing the equations:
1. First equation [tex]\( y = 2x - 1 \)[/tex]:
- The y-intercept is -1, so the line crosses the y-axis at -1.
- The slope is 2, which means for every 1 unit increase in x, y increases by 2 units.
- Points to plot: (0, -1), (1, 1), (2, 3), etc.
2. Second equation [tex]\( y = -\frac{1}{3}x + \frac{4}{3} \)[/tex]:
- The y-intercept is [tex]\( \frac{4}{3} \approx 1.33 \)[/tex].
- The slope is [tex]\( -\frac{1}{3} \)[/tex], which means for every 1 unit increase in x, y decreases by [tex]\( \frac{1}{3} \)[/tex] units.
- Points to plot: (0, [tex]\( \frac{4}{3} \)[/tex]), (3, 1), (-3, 2), etc.
### Intersection Point:
The intersection point of the two lines can be solved algebraically or visually from the graph. After solving, we find that:
[tex]\[ x = 1, \quad y = 1 \][/tex]
Therefore, the correct graph will show two lines intersecting at the point (1, 1). The graph should display:
- A line passing through points such as (0, -1) and (1, 1) for the first equation.
- Another line passing through points such as (0, [tex]\( \frac{4}{3} \)[/tex]) and (3, 1) for the second equation.
Choose the graph that matches these characteristics, with the intersection point clearly at (1, 1).
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