Answer: (x - a) is equal to (y + b).
To solve the given equations using the cross-multiplication method, we will start by multiplying the first equation by b and the second equation by y to eliminate the y term.
Multiplying the first equation by b gives us:
b(x - y) = b(a + b)
Multiplying the second equation by y gives us:
y(ax + by) = y(a^2 - b^2)
Expanding both equations, we get:
bx - by = ab + b^2 (Equation 1)
yax + yby = ya^2 - yb^2 (Equation 2)
Now, let's combine like terms:
In Equation 1, we have bx - by. We can factor out b:
b(x - y) = ab + b^2
In Equation 2, we have yax + yby. We can factor out y:
y(ax + by) = ya^2 - yb^2
After factoring, the equations become:
b(x - y) = b(a + b) (Equation 3)
y(ax + by) = y(a^2 - b^2) (Equation 4)
Now, let's cross multiply:
In Equation 3, we cross multiply to eliminate the fractions:
b(x - y) = b(a + b)
bx - by = ab + b^2
In Equation 4, we cross multiply to eliminate the fractions:
y(ax + by) = y(a^2 - b^2)
yax + yby = ya^2 - yb^2
Now, let's simplify the equations further:
In Equation 3, we have bx - by = ab + b^2. We can rearrange this equation:
bx - ab = by + b^2
In Equation 4, we have yax + yby = ya^2 - yb^2. We can rearrange this equation:
yax - ya^2 = -yby - yb^2
Now, let's factor out common terms:
In Equation 3, we can factor out b:
b(x - a) = b(y + b)
In Equation 4, we can factor out ya:
ya(x - a) = -yb(y + b)
Finally, we have two equations in the form of (x - a) = (y + b). By comparing the equations, we can see that both expressions in the parentheses are equal:
(x - a) = (y + b)
This is the solution using the cross-multiplication method. It shows that (x - a) is equal to (y + b).
Step-by-step explanation: