To find the solutions for the equation [tex]\(2x - y + al = 0\)[/tex], we need to manipulate the equation to express [tex]\(y\)[/tex] in terms of [tex]\(x\)[/tex], assuming [tex]\(a\)[/tex] and [tex]\(l\)[/tex] are constants:
1. Start with the equation:
[tex]\[
2x - y + al = 0
\][/tex]
2. Rearrange the equation to solve for [tex]\(y\)[/tex]:
\begin{align}
2x - y + al &= 0 \\
2x + al &= y \\
y &= 2x + al
\end{align}
3. Now we need to find two distinct solutions by choosing different values for [tex]\(x\)[/tex]. Let's choose [tex]\(x\)[/tex] values arbitrarily:
- For the first solution, choose [tex]\(x = 1\)[/tex].
[tex]\[
y = 2(1) + al
\][/tex]
Simplifying,
[tex]\[
y = 2 + al
\][/tex]
So, the first solution is:
[tex]\[
(x, y) = (1, 2 + al)
\][/tex]
- For the second solution, choose [tex]\(x = 2\)[/tex].
[tex]\[
y = 2(2) + al
\][/tex]
Simplifying,
[tex]\[
y = 4 + al
\][/tex]
So, the second solution is:
[tex]\[
(x, y) = (2, 4 + al)
\][/tex]
Given that [tex]\(a\)[/tex] and [tex]\(l\)[/tex] are constants, let's consider specifying [tex]\(a = 1\)[/tex] and [tex]\(l = 1\)[/tex] for these values to be specific:
- For [tex]\(x = 1\)[/tex]:
[tex]\[
y = 2 + 1 \cdot 1 = 3
\][/tex]
So the first specific solution is:
[tex]\[
(x, y) = (1, 3)
\][/tex]
- For [tex]\(x = 2\)[/tex]:
[tex]\[
y = 4 + 1 \cdot 1 = 5
\][/tex]
So the second specific solution is:
[tex]\[
(x, y) = (2, 5)
\][/tex]
Therefore, the two solutions to the equation [tex]\(2x - y + al = 0\)[/tex] are [tex]\((1, 3)\)[/tex] and [tex]\((2, 5)\)[/tex].
