Question 215929
Find a number that is 68 greater than three times its opposite.<br>
Let x = "a number".
then the opposite of that number = -x
and three times the opposite = 3(-x) = -3x
Now we do a "translation" of "a number is 68 greater than three times its opposite". When doing a translation from English to Algebra, forms of the verb "to be" (is, are, was, were, etc.) translate into equal signs.
"a number is 68 greater than three times its opposite"
    x     =  68    +               (-3x)
Now we have an equation we can solve. As usual there are many ways to solve this. One way would be: Add 3x to both sides:
x + 3x = 68 + (-3x) + 3x
Add like terms:
4x = 68 + 0
4x = 68
Divide both sides by 4:
{{{(4x)/4 = (68)/4}}}
Cancel common factors
{{{(4*x)/(4*1) = (17*4)/(4*1)}}}
{{{(cross(4)*x)/(cross(4)*1) = (17*cross(4))/(cross(4)*1)}}}
leaving
{{{x/1 = 17/1}}}
Simplifying we get:
{{{x = 17}}}