Smith Chan
08/08/2023 · Junior High School
b) \( \left\{\begin{array}{c}x+y=6 \\ 2 x-b y=4\end{array}\right. \)
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Tutor-Verified Answer
Step-by-step Solution
To solve the system of equations
\[
\begin{cases}
x + y = 6 \\
2x - by = 4
\end{cases}
\]
we can start by expressing one variable in terms of the other using the first equation. From the first equation, we can express \(y\) in terms of \(x\):
\[
y = 6 - x
\]
Now, we can substitute this expression for \(y\) into the second equation:
\[
2x - b(6 - x) = 4
\]
Expanding this gives:
\[
2x - 6b + bx = 4
\]
Now, we can combine like terms:
\[
(2 + b)x - 6b = 4
\]
Next, we can isolate \(x\):
\[
(2 + b)x = 4 + 6b
\]
\[
x = \frac{4 + 6b}{2 + b}
\]
Now that we have \(x\), we can substitute it back into the equation for \(y\):
\[
y = 6 - x = 6 - \frac{4 + 6b}{2 + b}
\]
To simplify \(y\), we can find a common denominator:
\[
y = \frac{(6(2 + b) - (4 + 6b))}{2 + b}
\]
Calculating the numerator:
\[
6(2 + b) = 12 + 6b
\]
So,
\[
y = \frac{12 + 6b - 4 - 6b}{2 + b} = \frac{8}{2 + b}
\]
Thus, the solution to the system of equations is:
\[
\begin{cases}
x = \frac{4 + 6b}{2 + b} \\
y = \frac{8}{2 + b}
\end{cases}
\]
This gives us the values of \(x\) and \(y\) in terms of the parameter \(b\).
Quick Answer
The solution to the system of equations is:
\[
\begin{cases}
x = \frac{4 + 6b}{2 + b} \\
y = \frac{8}{2 + b}
\end{cases}
\]
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