# If y varies jointly with x and z, and y = 24 when x = 3 and z = 2, then the value of x when y = 6...

## Question:

If {eq}y {/eq} varies jointly with {eq}x {/eq} and {eq}z {/eq}, and {eq}y = 24 {/eq} when {eq}x = 3 {/eq} and {eq}z = 2 {/eq}, then the value of {eq}x {/eq} when {eq}y = 6 {/eq} and {eq}z = 0.5 {/eq} is {eq}\rule{1cm}{0.15mm} {/eq}.

## Joint Variation:

If a variable {eq}c {/eq} varies jointly with respect to both the variables {eq}a {/eq} and {eq}b {/eq}, then it means that:

$$c= kab$$

where, {eq}k {/eq} is a constant of variation.

Note: It means that {eq}c {/eq} is directly proportional to each of {eq}a {/eq} and {eq}b {/eq} taken one at a time.

The problem says, "{eq}y {/eq} varies jointly with {eq}x {/eq} and {eq}z {/eq}.

So by the definition of joint variation,

$$y=kxz \,\,\,\,\,\,\rightarrow (1)$$

Here, {eq}k {/eq} is a proportionality constant.

Substitute the first set of values {eq}x=3; \, y=24; \, z=2 {/eq} in (1) and solve for {eq}k {/eq}:

$$24= k(3)(2) \\ 24=6k \\ \text{Dividing both sides by 6}, \\ k=4$$

Substitute this and second set of given values {eq}y=6; \,\, z=0.5 {/eq} in (1) and solve for {eq}x {/eq}:

$$6= 4(x)(0.5) \\ 6=2x \\ \text{Dividing both sides by 2}, \\ x= \boxed{\mathbf{3}}$$