# Determine x,y,z if x/2=y/3=z/5 and x+y+z=20

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### 3 Answers

Given that:

x/2 = y/3 = z/5..........(1)

x+ y + z = 20.........(2)

From (1) we will rewrite the numbers as functions of x.

==> x/2 = y/3

==> y= (3/2)x ............(3)

==> x/2 = z/5

==> z = (5/2)x .............(4)

Now we will substitute (3) and (4) into (2).

==> x + y + z = 20

==> x + (3/2)x + (5/2)x = 20

==> (2+ 3+5)/2 x = 20

==> 10/2 x = 20

==> 5x = 20

==> x = 20/5 = 4

==> y= (3/2)x = 3/2 * 4 = 6

==> z= (5/2)x = 5/2 * 4 = 10

**Then the numbers are: ( x, y, z) = ( 4, 6, 10)**

We have to determine x,y,z given that x/2=y/3=z/5 and x+y+z=20

x/2=y/3=z/5

=> y = 3x/2 and z = 5x/2

Substitute in x + y + z = 20

=> x + 3x/2 + 5x/2 = 20

=> 2x + 3x + 5x = 40

=> 10x = 40

=> x = 4

y = 6

z = 10

**The value of x = 4, y = 6 and z = 10**

We'll express the y and z terms with respect to x.

x/2 = y/3 <=> y = 3x/2

x/2 = z/5 <=> z = 5x/2

x+y+z=20 <=> x + 3x/2 + 5x/2 = 20

x + 8x/2 = 20

x + 4x = 20

5x = 20 => x = 4

Since x = 4 => y = 3*4/2 => y = 6

x = 4 => z = 5*4/2 => z = 10

**The required values of (x , y , z) are: (4 , 6 , 10).**