Question

Solve for x; (a+bx)/(a+b)=(c+dx)/(c+d) if cb=ad

Answer 1

To solve for x, we can start by cross-multiplying the equation to eliminate the denominators:

(a+bx)(c+d) = (c+dx)(a+b)

Expanding the terms on both sides:

ac + adx + bc + bdx^2 = ac + abx + cdx + bd

Simplifying and rearranging the terms:

adx + bdx^2 - abx - cdx = bd - ac

dx(ad - ab - c) = bd - ac

Now, since we know that cb=ad, we can substitute ad=cb into the equation:

dx(cb - ab - c) = bd - ac

dx(cb - ab - c) = b(cd - ac)

x = b(cd - ac)/(d(cb - ab - c))

Therefore, the solution for x is:

x = b(cd - ac)/(d(cb - ab - c))

Answer 2

So the solution is that x can be any real number if ( bd = ab) , otherwise (x = 0 ).

To solve for x in the equation

`\[ (a + bx)/(a + b) = (c + dx)/(c + d) \]`

given that cb = ad , we can cross-multiply to eliminate the fractions:

`\[ (a + bx)(c + d) = (c + dx)(a + b) \]`

Expanding both sides:

`\[ ac + adx + bcx + bdx = ac + abx + cdx + cbd \]`

Using the given condition cb = ad , we can simplify the equation:

`\[ ac + adx + adx + bdx = ac + abx + cdx + adx \]`

Combine like terms:

`\[ ac + 2adx + bdx = ac + abx + cdx + adx \]`

Since ac appears on both sides of the equation, we can cancel it out:

`\[ 2adx + bdx = abx + cdx + adx \]`

Now we have an equation with terms containing x on both sides. Since we want to solve for x , we can gather all terms involving x on one side of the equation:

`\[ 2adx + bdx - abx - cdx - adx = 0 \]`

Simplify by combining like terms:

`\[ adx + bdx - abx - cdx = 0 \]`

Now, we factor out an x from each term:

`\[ x(ad + bd - ab - cd) = 0 \]`

To solve for x , we can divide both sides by the factor
`\( (ad + bd - ab - cd) \),` provided that this factor is not equal to zero. However, since we have cb = ad , the equation simplifies to:

`\[ x(ad + bd - ab - cb) = 0 \]`

`\[ x(ad + bd - ab - ad) = 0 \]`

`\[ x(bd - ab) = 0 \]`

Now, if
`\( bd - ab = 0 \),` then we would have x times 0 which gives us no information about x (since any number times 0 equals 0). Therefore, if
`\( bd = ab \),` x can be any real number. If
`\( bd \\eq ab \),` then x must be 0 to satisfy the equation.