How much water should be mixed with 237 ml of ammonia, whose strength is 100%, in order to create a mixture that is diluted to a 75% strength.

Let the x ml of water be mixed with 237 ml of ammonia whose strength is 100%,  in order to create a mixture of 75% strength.
The equation required to solve this will be:
237/(237+x)=75/100
solving for x we get:
237/(237+x)=3/4
237×4=3(237+x)
948=711+3x
3x+711=948
3x=948-711
3x=237
x=237/3
x=79 ml
therefore the amount of water to be added will be 79 ml

Let the amount of water to be mixed = x ml 
new mixture = 237+x 
The new mixture is diluted to 75% strength i.e. 75/100 
75/100 = 3/4 
Therefore, 
237/(237+x) = 3/4 
4 *237 =3 (237+x) 
948=711+3x 
3x=948-711 
3x=237 
x=237/3 
x=79 
The amount of water to be mixed with ammonia = 79ml

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