Sure! Let's complete the table step by step.
Given equations are:
1. [tex]\(3x + 4y = 12\)[/tex]
2. [tex]\(3x + 4y = 4\)[/tex]
3. [tex]\(3x + 4y = 3\)[/tex]
4. [tex]\(3x + 4y = 2\)[/tex]
To find the [tex]\(x\)[/tex]-intercept, we set [tex]\(y = 0\)[/tex] and solve for [tex]\(x\)[/tex]. Similarly, to find the [tex]\(y\)[/tex]-intercept, we set [tex]\(x = 0\)[/tex] and solve for [tex]\(y\)[/tex].
### For the equation [tex]\(3x + 4y = 12\)[/tex]:
- [tex]\(x\)[/tex]-intercept:
- Set [tex]\(y = 0\)[/tex]:
[tex]\[
3x + 4(0) = 12 \implies 3x = 12 \implies x = 4
\][/tex]
- [tex]\(y\)[/tex]-intercept:
- Set [tex]\(x = 0\)[/tex]:
[tex]\[
3(0) + 4y = 12 \implies 4y = 12 \implies y = 3
\][/tex]
### For the equation [tex]\(3x + 4y = 4\)[/tex]:
- [tex]\(x\)[/tex]-intercept:
- Set [tex]\(y = 0\)[/tex]:
[tex]\[
3x + 4(0) = 4 \implies 3x = 4 \implies x = \frac{4}{3}
\][/tex]
### For the equation [tex]\(3x + 4y = 3\)[/tex]:
- [tex]\(x\)[/tex]-intercept:
- Set [tex]\(y = 0\)[/tex]:
[tex]\[
3x + 4(0) = 3 \implies 3x = 3 \implies x = 1
\][/tex]
### For the equation [tex]\(3x + 4y = 2\)[/tex]:
- [tex]\(x\)[/tex]-intercept:
- Set [tex]\(y = 0\)[/tex]:
[tex]\[
3x + 4(0) = 2 \implies 3x = 2 \implies x = \frac{2}{3}
\][/tex]
- [tex]\(y\)[/tex]-intercept:
- Set [tex]\(x = 0\)[/tex]:
[tex]\[
3(0) + 4y = 2 \implies 4y = 2 \implies y = \frac{1}{2}
\][/tex]
Now, let’s fill in the table with these results:
[tex]\[
\begin{array}{|ccc|}
\hline
\text{Equations} & \text{x-intercept} & \text{y-intercept} \\
\hline
3x + 4y = 12 & 4 & 3 \\
3x + 4y = 4 & \frac{4}{3} & \\
3x + 4y = 3 & 1 & \\
3x + 4y = 2 & \frac{2}{3} & \frac{1}{2} \\
\hline
\end{array}
\][/tex]
Thus, the complete table is:
[tex]\[
\begin{tabular}{|ccc|}
\hline
\text{Equations} & \text{x-intercept} & \text{y-intercept} \\
\hline
3x + 4y = 12 & 4 & 3 \\
3x + 4y = 4 & \frac{4}{3} & \\
3x + 4y = 3 & 1 & \\
3x + 4y = 2 & \frac{2}{3} & \frac{1}{2} \\
\hline
\end{tabular}
\][/tex]
