What number could replace

below?`SYMBOL`

`\dfrac{`

`A`}{`B`} = \dfrac{`C`}{`SYMBOL`}

To get the right numerator

, the left numerator `C`

is multiplied by `A`

.`M`

To find the right denominator, multiply the left denominator by

as well.`M`

`B` \times `M` = `D`

Notice both the numerator and denominator are being multiplied by `{`

.`M`}

We can write that as `\dfrac{`

, which is equal to `M`}{`M`}`1`

when reduced.

So we can solve this problem by multiplying the fraction on the left by `1`

.

The equation becomes: `\dfrac{`

so our answer is `A`}{`B`} \times \dfrac{`M`}{`M`} = \dfrac{`C`}{`D`}

.`D`

What number could replace

below?`SYMBOL`

`\dfrac{`

`A`}{`B`} = \dfrac{`SYMBOL`}{`D`}

To get the right denominator

, the left denominator `D`

is multiplied by `B`

.`M`

To find the right numerator, multiply the left numerator by

as well.`M`

`A` \times `M` = `C`

Notice both the numerator and denominator are being multiplied by `{`

.`M`}

We can write that as `\dfrac{`

, which is equal to `M`}{`M`}`1`

when reduced.

So we can solve this problem by multiplying the fraction on the left by `1`

.

The equation becomes: `\dfrac{`

so our answer is `A`}{`B`} \times \dfrac{`M`}{`M`} = \dfrac{`C`}{`D`}

.`C`

What number could replace

below?`SYMBOL`

`\dfrac{`

`C`}{`D`} = \dfrac{`A`}{`SYMBOL`}

To get the right numerator

, the left numerator `A`

is divided by `C`

.`M`

To find the right denominator, divide the left denominator by

as well.`M`

`D` \div `M` = `B`

Notice both the numerator and denominator are being divided by `{`

.`M`}

We can write that as `\dfrac{`

, which is equal to `M`}{`M`}`1`

when reduced.

So we can solve this problem by dividing the fraction on the left by `1`

.

The equation becomes: `\dfrac{`

so our answer is `C`}{`D`} \div \dfrac{`M`}{`M`} = \dfrac{`A`}{`B`}

.`B`

What number could replace

below?`SYMBOL`

`\dfrac{`

`C`}{`D`} = \dfrac{`SYMBOL`}{`B`}

To get the right denominator

, the left denominator `B`

is divided by `D`

.`M`

To find the right numerator, divide the left numerator by

as well.`M`

`C` \div `M` = `A`

Notice both the numerator and denominator are being divided by `{`

.`M`}

We can write that as `\dfrac{`

, which is equal to `M`}{`M`}`1`

when reduced.

So we can solve this problem by dividing the fraction on the left by `1`

.

The equation becomes: `\dfrac{`

so our answer is `C`}{`D`} \div \dfrac{`M`}{`M`} = \dfrac{`A`}{`B`}

.`A`