# solve the following pair of simultaneous equation:

(x/a)+(y/b)=1

(x/b)+(y/a)=1

can someone help please?i have no idea how to solve it =(

x/(ab) + y/b^2 = 1/b

x/(ab) + y/a^2 = 1/a

Subtract one from the other

y(1/b^2 - 1/a^2) = 1/b - 1/a

y(1/b + 1/a)(1/b - 1/a) = (1/b - 1/a)

Divide both sides by (1/b - 1/a)

y = 1/(1/a + 1/b)= ab/(a+b)

Now substitute this y into an earlier equation solve for x. I get x = y

## That's because x=y is the answer...

## Well, well, well. It seems we have quite the equation conundrum here. Fear not, dear friend, for Clown Bot is here to help! Let's tackle this simultaneous equation with a twist of humor, shall we?

So we've got these two equations:

(x/a) + (y/b) = 1

(x/b) + (y/a) = 1

Now, to solve this, we can use some math magic and a dash of clownery. Let's start by rearranging the second equation to solve for x:

x = 1 - (y/a)

Now, let's substitute this value of x into the first equation:

(1 - (y/a))/a + (y/b) = 1

Now, let's multiply everything by ab to get rid of those pesky fractions:

a(b - y) + b(y) = ab

Let's simplify:

ab - ay + by = ab

Now, let's collect like terms:

by - ay = 0

And factor out the y:

y(b-a) = 0

Now, there are two possibilities: either y = 0, or (b - a) = 0.

If y = 0, then we can substitute y = 0 into either equation to solve for x. So, let's say y = 0.

x/a + 0/b = 1

x/a = 1

This means x = a.

If (b - a) = 0, then we have b = a. In this case, both equations are essentially the same equation repeated twice, so any value of x and y that satisfies one equation will satisfy the other. So we can't determine a unique solution in this case.

So, in conclusion, we have two possible solutions:

1) x = a, y = 0

2) If b = a, then there are infinitely many solutions.

I hope that brings a smile to your face, my friend! Mathematics may be serious, but solving equations can still be a clowning good time.

## First, let's rewrite the given equations:

Equation 1: (x/a) + (y/b) = 1

Equation 2: (x/b) + (y/a) = 1

To solve this system of equations, we can use the method of substitution.

Step 1: Solve one equation for one variable in terms of the other variable.

Let's solve Equation 1 for x:

x = a - (y/b)

Step 2: Substitute x into Equation 2.

Replace x in Equation 2 with the expression we obtained from step 1:

(a - (y/b))/b + (y/a) = 1

Now, let's simplify this equation.

(a/b) - (y/b^2) + (y/a) = 1

Multiply through by the common denominator, ab.

(a^2 - y + by)/ab = 1

Multiply both sides of the equation by ab to eliminate the fraction.

a^2 - y + by = ab

Step 3: Solve for y.

Rearrange the equation to isolate y.

by - y = ab - a^2

y = ab - a^2 + by

Group the y terms together.

y - by = ab - a^2

Factor out y from the terms.

y(1 - b) = ab - a^2

Divide both sides by (1 - b).

y = (ab - a^2)/(1 - b)

Step 4: Solve for x.

Now that we have the value of y, we can substitute it back into either of the original equations to solve for x. Let's use Equation 1.

(x/a) + ((ab - a^2)/(1 - b))/b = 1

Multiply through by ab(1 - b) to eliminate the fractions.

(x(1 - b) + (ab - a^2))/a = ab(1 - b)

Distribute x.

x - bx + ab - a^2 = ab - ab^2

Rearrange the equation.

x - bx - ab + a^2 = -ab^2

Combine like terms.

x(1 - b) = -ab^2

Divide both sides by (1 - b).

x = -ab^2/(1 - b)

So, the solution to the pair of simultaneous equations is:

x = -ab^2/(1 - b)

y = (ab - a^2)/(1 - b)

## To solve the pair of simultaneous equations:

1. Start with the equations given:

(x/a) + (y/b) = 1

(x/b) + (y/a) = 1

2. Multiply both sides of the first equation by b, and multiply both sides of the second equation by a to eliminate the fractions. This gives:

bx/a + by/b = b

ax/b + ay/a = a

3. Simplify both equations:

bx/a + y = b

x + ay/b = a

4. Rearrange the equations to isolate x and y:

bx/a = b - y, solve for x: x = (a/b) * (b - y)

x = a - (ay/b), solve for y: y = (ab)/(a + b)

5. Substitute the value of y into the equation for x:

x = a - (a * (ab)/(a + b))

= a - (a^2 * b)/(a + b)

So the solution to the simultaneous equations is:

x = a - (a^2 * b)/(a + b)

y = (ab)/(a + b)