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Sagot :
Let's solve the problem step-by-step.
### Step 1: Create T-charts for both equations
#### Equation 1: [tex]\( y = 3x - 15 \)[/tex]
To create a T-chart, we substitute different values of [tex]\( x \)[/tex] and compute the corresponding values of [tex]\( y \)[/tex]:
| [tex]\( x \)[/tex] | [tex]\( y = 3x - 15 \)[/tex] | [tex]\( (x, y) \)[/tex] |
|--------|-------------------|------------|
| 0 | [tex]\( 3(0) - 15 = -15 \)[/tex] | (0, -15) |
| 1 | [tex]\( 3(1) - 15 = -12 \)[/tex] | (1, -12) |
| 2 | [tex]\( 3(2) - 15 = -9 \)[/tex] | (2, -9) |
| 3 | [tex]\( 3(3) - 15 = -6 \)[/tex] | (3, -6) |
| 4 | [tex]\( 3(4) - 15 = -3 \)[/tex] | (4, -3) |
We have the points: [tex]\((0, -15)\)[/tex], [tex]\((1, -12)\)[/tex], [tex]\((2, -9)\)[/tex], [tex]\((3, -6)\)[/tex], [tex]\((4, -3)\)[/tex].
#### Equation 2: [tex]\( x + y = 13 \)[/tex]
Similarly, substituting different values of [tex]\( x \)[/tex] and computing [tex]\( y \)[/tex]:
| [tex]\( x \)[/tex] | [tex]\( y = 13 - x \)[/tex] | [tex]\( (x, y) \)[/tex] |
|--------|-------------------|------------|
| 0 | [tex]\( 13 - 0 = 13 \)[/tex] | (0, 13) |
| 1 | [tex]\( 13 - 1 = 12 \)[/tex] | (1, 12) |
| 2 | [tex]\( 13 - 2 = 11 \)[/tex] | (2, 11) |
| 3 | [tex]\( 13 - 3 = 10 \)[/tex] | (3, 10) |
| 4 | [tex]\( 13 - 4 = 9 \)[/tex] | (4, 9) |
We have the points: [tex]\((0, 13)\)[/tex], [tex]\((1, 12)\)[/tex], [tex]\((2, 11)\)[/tex], [tex]\((3, 10)\)[/tex], [tex]\((4, 9)\)[/tex].
### Step 2: Graph the points and find the intersection
Plotting these points on a graph, we can draw the lines for the equations:
- For [tex]\( y = 3x - 15 \)[/tex], draw a line passing through the points [tex]\((0, -15)\)[/tex], [tex]\((1, -12)\)[/tex], [tex]\((2, -9)\)[/tex], [tex]\((3, -6)\)[/tex], [tex]\((4, -3)\)[/tex].
- For [tex]\( x + y = 13 \)[/tex], draw a line passing through the points [tex]\((0, 13)\)[/tex], [tex]\((1, 12)\)[/tex], [tex]\((2, 11)\)[/tex], [tex]\((3, 10)\)[/tex], [tex]\((4, 9)\)[/tex].
### Step 3: Find the solution at the intersection
The solution to the system of equations is where these two lines intersect. By examining our coordinates and graph, we find that the intersection of the two lines occurs at the point [tex]\((7, 6)\)[/tex].
### Summary
- T-chart for [tex]\( y = 3x - 15\)[/tex]: [tex]\((0, -15)\)[/tex], [tex]\((1, -12)\)[/tex], [tex]\((2, -9)\)[/tex], [tex]\((3, -6)\)[/tex], [tex]\((4, -3)\)[/tex].
- T-chart for [tex]\( x + y = 13\)[/tex]: [tex]\((0, 13)\)[/tex], [tex]\((1, 12)\)[/tex], [tex]\((2, 11)\)[/tex], [tex]\((3, 10)\)[/tex], [tex]\((4, 9)\)[/tex].
- The intersection point (solution) is [tex]\((7, 6)\)[/tex].
Thus, the solution to the system of equations is [tex]\((7, 6)\)[/tex].
### Step 1: Create T-charts for both equations
#### Equation 1: [tex]\( y = 3x - 15 \)[/tex]
To create a T-chart, we substitute different values of [tex]\( x \)[/tex] and compute the corresponding values of [tex]\( y \)[/tex]:
| [tex]\( x \)[/tex] | [tex]\( y = 3x - 15 \)[/tex] | [tex]\( (x, y) \)[/tex] |
|--------|-------------------|------------|
| 0 | [tex]\( 3(0) - 15 = -15 \)[/tex] | (0, -15) |
| 1 | [tex]\( 3(1) - 15 = -12 \)[/tex] | (1, -12) |
| 2 | [tex]\( 3(2) - 15 = -9 \)[/tex] | (2, -9) |
| 3 | [tex]\( 3(3) - 15 = -6 \)[/tex] | (3, -6) |
| 4 | [tex]\( 3(4) - 15 = -3 \)[/tex] | (4, -3) |
We have the points: [tex]\((0, -15)\)[/tex], [tex]\((1, -12)\)[/tex], [tex]\((2, -9)\)[/tex], [tex]\((3, -6)\)[/tex], [tex]\((4, -3)\)[/tex].
#### Equation 2: [tex]\( x + y = 13 \)[/tex]
Similarly, substituting different values of [tex]\( x \)[/tex] and computing [tex]\( y \)[/tex]:
| [tex]\( x \)[/tex] | [tex]\( y = 13 - x \)[/tex] | [tex]\( (x, y) \)[/tex] |
|--------|-------------------|------------|
| 0 | [tex]\( 13 - 0 = 13 \)[/tex] | (0, 13) |
| 1 | [tex]\( 13 - 1 = 12 \)[/tex] | (1, 12) |
| 2 | [tex]\( 13 - 2 = 11 \)[/tex] | (2, 11) |
| 3 | [tex]\( 13 - 3 = 10 \)[/tex] | (3, 10) |
| 4 | [tex]\( 13 - 4 = 9 \)[/tex] | (4, 9) |
We have the points: [tex]\((0, 13)\)[/tex], [tex]\((1, 12)\)[/tex], [tex]\((2, 11)\)[/tex], [tex]\((3, 10)\)[/tex], [tex]\((4, 9)\)[/tex].
### Step 2: Graph the points and find the intersection
Plotting these points on a graph, we can draw the lines for the equations:
- For [tex]\( y = 3x - 15 \)[/tex], draw a line passing through the points [tex]\((0, -15)\)[/tex], [tex]\((1, -12)\)[/tex], [tex]\((2, -9)\)[/tex], [tex]\((3, -6)\)[/tex], [tex]\((4, -3)\)[/tex].
- For [tex]\( x + y = 13 \)[/tex], draw a line passing through the points [tex]\((0, 13)\)[/tex], [tex]\((1, 12)\)[/tex], [tex]\((2, 11)\)[/tex], [tex]\((3, 10)\)[/tex], [tex]\((4, 9)\)[/tex].
### Step 3: Find the solution at the intersection
The solution to the system of equations is where these two lines intersect. By examining our coordinates and graph, we find that the intersection of the two lines occurs at the point [tex]\((7, 6)\)[/tex].
### Summary
- T-chart for [tex]\( y = 3x - 15\)[/tex]: [tex]\((0, -15)\)[/tex], [tex]\((1, -12)\)[/tex], [tex]\((2, -9)\)[/tex], [tex]\((3, -6)\)[/tex], [tex]\((4, -3)\)[/tex].
- T-chart for [tex]\( x + y = 13\)[/tex]: [tex]\((0, 13)\)[/tex], [tex]\((1, 12)\)[/tex], [tex]\((2, 11)\)[/tex], [tex]\((3, 10)\)[/tex], [tex]\((4, 9)\)[/tex].
- The intersection point (solution) is [tex]\((7, 6)\)[/tex].
Thus, the solution to the system of equations is [tex]\((7, 6)\)[/tex].
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