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X-y=a+b,ax+by=a2-b2 using cross multiplication method

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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:

User James Bucanek
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