Answer:
To solve this system of equations by elimination, we can first multiply the two equations by necessary multiples such that the coefficients of y's in both equations are the same:
1. Multiply the first equation by b:
bx - by = ab + b^2
2. Multiply the second equation by 1:
ax + by = a^2 - b^2
Now, add both equations to eliminate the y variable:
bx - by + ax + by = ab + b^2 + a^2 - b^2
Combine like terms:
(a + b)x = a^2 + ab
Now, divide by (a + b) to solve for x:
x = (a^2 + ab) / (a + b)
Now that we have the value of x, substitute it into one of the original equations to solve for y. Let's use the first equation:
x - y = a + b
Substitute the expression for x:
((a^2 + ab) / (a + b)) - y = a + b
Simplify and solve for y.
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Answer:
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Step-by-step explanation:
To solve the system of equations by elimination:
1. \( x - y = a + b \)
2. \( ax + by = a^2 - b^2 \)
First, let's rewrite the equations clearly:
1. \( x - y = a + b \) \(\quad \text{(Equation 1)}\)
2. \( ax + by = a^2 - b^2 \) \(\quad \text{(Equation 2)}\)
We aim to eliminate one of the variables. Let’s start by eliminating \( y \). To do this, we can multiply Equation 1 by \( b \) and Equation 2 by 1 (so it stays the same):
\[
b(x - y) = b(a + b) \implies bx - by = ab + b^2 \quad \text{(Equation 3)}
\]
We now have:
\[
bx - by = ab + b^2 \quad \text{(Equation 3)}
\]
\[
ax + by = a^2 - b^2 \quad \text{(Equation 2)}
\]
Next, we add Equation 3 and Equation 2:
\[
bx - by + ax + by = ab + b^2 + a^2 - b^2
\]
The terms \( -by \) and \( +by \) cancel out:
\[
bx + ax = ab + a^2
\]
Combine the like terms:
\[
(b + a)x = ab + a^2
\]
Solve for \( x \):
\[
x = \frac{ab + a^2}{a + b}
\]
Now we substitute \( x \) back into Equation 1 to solve for \( y \):
\[
x - y = a + b
\]
\[
\frac{ab + a^2}{a + b} - y = a + b
\]
Rearrange to solve for \( y \):
\[
\frac{ab + a^2}{a + b} - (a + b) = y
\]
To simplify:
\[
y = \frac{ab + a^2 - (a + b)^2}{a + b}
\]
Expanding \((a + b)^2\):
\[
(a + b)^2 = a^2 + 2ab + b^2
\]
Thus:
\[
y = \frac{ab + a^2 - a^2 - 2ab - b^2}{a + b}
\]
\[
y = \frac{ab + a^2 - a^2 - 2ab - b^2}{a + b}
\]
\[
y = \frac{ab + a^2 - a^2 - 2ab - b^2}{a + b}
\]
\[
y = \frac{-ab - b^2}{a + b}
\]
\[
y = -\frac{ab + b^2}{a + b}
\]
\[
y = -\frac{b(a + b)}{a + b}
\]
Since \(a + b \neq 0\) (we assume \(a + b \neq 0\)), we can cancel \(a + b\):
\[
y = -b
\]
Therefore, the solutions to the system of equations are:
\[
x = a, \quad y = -b
\]
