To solve the system of equations [tex]\(2x + y = 4\)[/tex] and [tex]\(x + y = 2\)[/tex] simultaneously, follow these steps:
1. Start with the two given equations:
[tex]\[
2x + y = 4 \quad \text{(Equation 1)}
\][/tex]
[tex]\[
x + y = 2 \quad \text{(Equation 2)}
\][/tex]
2. Solve for [tex]\(y\)[/tex] from Equation 2:
[tex]\[
y = 2 - x
\][/tex]
3. Substitute [tex]\(y = 2 - x\)[/tex] into Equation 1 to eliminate [tex]\(y\)[/tex]:
[tex]\[
2x + (2 - x) = 4
\][/tex]
Simplify the equation:
[tex]\[
2x + 2 - x = 4
\][/tex]
[tex]\[
x + 2 = 4
\][/tex]
[tex]\[
x = 4 - 2
\][/tex]
[tex]\[
x = 2
\][/tex]
4. Substitute [tex]\(x = 2\)[/tex] back into Equation 2 to find [tex]\(y\)[/tex]:
[tex]\[
x + y = 2
\][/tex]
[tex]\[
2 + y = 2
\][/tex]
[tex]\[
y = 2 - 2
\][/tex]
[tex]\[
y = 0
\][/tex]
Therefore, the solution to the system of equations is:
[tex]\[
x = 2, \quad y = 0
\][/tex]
For the table, the relation provided is [tex]\(y = 2x + 5\)[/tex]:
To answer question 6:
Use [tex]\(x = 3\)[/tex]:
[tex]\[
y = 2 \cdot 3 + 5 = 6 + 5 = 11
\][/tex]
So, [tex]\( A = 11 \)[/tex].
To answer question 7:
Use [tex]\(x = 1\)[/tex]:
[tex]\[
y = 2 \cdot 1 + 5 = 2 + 5 = 7
\][/tex]
So, [tex]\( c = 7 \)[/tex].
Update the table:
[tex]\[
\begin{tabular}{|l|l|l|l|l|l|l|l|l|l|}
\hline
$x$ & 4 & 3 & 2 & 2 & 0 & 1 & 2 & 3 & 4 \\
\hline
$\times$ & 3 & 11 & 1 & 8 & 5 & 7 & 9 & 0 & 13 \\
\hline
\end{tabular}
\][/tex]
