1. ## Solve the following.

1. 4/x = 5/7

2. 3a/7 = -2/5

3. x+1/6 = 4/3

4. 24/x-3 = 72/x+3

Thanks.

2. I'll only solve the 1st one, I think you can do the rest.

$\frac{4}{x} = \frac{5}{7}$

Multiply both sides by $x$,

$\frac{4}{x}\cdot x = \frac{5}{7}\cdot x$

$\frac{4}{\not x}\cdot \not x = \frac{5x}{7}$

$4 = \frac{5x}{7}$

Now multiply both sides by $7$,

$4\cdot 7 = \frac{5x}{7}\cdot 7$

$28 = \frac{5x}{\not 7}\cdot \not7$

$28 = 5x$

$x = \frac{28}{5}$

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If you have an equation in the from $\frac{a}{b} = \frac{c}{d}$, products of the terms crosswise are equal, $a\cdot d = b \cdot c$

So, $\frac{4}{x} = \frac{5}{7}$

$4\cdot 7 = 5\cdot x$

$5x = 28$

$x = \frac{28}{5}$

3. Originally Posted by elliotfsl
1. 4/x = 5/7
Initial Equation
$\frac 4x = \frac 57$

You want to get x in the numerator so multiply both sides by x
$\frac x1 * \frac 4x = \frac 57*\frac x1$

Simplifly
$4 = \frac {5x}7$

You want x to be by itself, so you need to get rid of the 7 in the denominator. You can do this by multiplying both sides by 7 since 7/7 =1
$\frac 71 *\frac 41 = \frac {5x}7*\frac 71$

Simplify
$28 = 5x$

You want to get x by itself, so you need to get rid of the 5, do this by multiplying both sides by 1/5 since 5/5=1
$\frac 15*28 = 5x\frac 15$

Simplify
$\frac{28}5 = x$

4. Originally Posted by elliotfsl
4. 24/x-3 = 72/x+3
Initial equation
$\frac {24}{x-3} = \frac{72}{x+3}$

Again, you need to get x out of the denominator, so multiply both sides by x-3
$24 = \frac{72(x-3)}{x+3}$

X is still in the denominator on the right side, so multiply both sides by x+3
$24(x+3) = 72(x-3)$

Distribute the coefficients
$24x+72 = 72x-216$

You need all your x's to be on the same side, so subtract 24x
$72 = 72x-24x-216$

Simplify
$72 = 48x-216$

You need all the other numbers to be on the other side of x, so add 216
$72 +216= 48x$

Simplify
$288= 48x$

You need x to be by itself, so multiply by 1/48
$\frac{288}{48}= x$

And simplify
$6 = x$