Question 943201
<pre>
He can check by substituting (1) for x.

3(3x+6) = 4(7x&#8722;1) 
3(3·1+6) = 4(7·1&#8722;1)
3(3+6) = 4(7-1)
3(9) = 4(6)
27 = 24

False so there is a mistake.


3(3x+6) = 4(7x&#8722;1) 
9x+18 = 28x&#8722;1        <--should be   9x+18 = 28x&#8722;4
9x&#8722;9x+18 = 28x&#8722;9x&#8722;1  <--should be   9x&#8722;9x+18 = 28x&#8722;9x&#8722;4
18 = 19x&#8722;1           <--should be   18 = 19x-4
18+1 = 19x&#8722;1+1       <--dhould be   18+4 = 19x-4+4 
19 = 19x             <--should be   22 = 19x 
1 = x                <--should be   {{{22/19}}} = x

Common error in distributing: A(B + C), forgetting to 
multiply the A by the C.

Check the terribly ugly answer {{{22/19}}}

{{{3(3x+6)}}}{{{""=""}}}{{{4(7x-1)}}}

{{{3(3(22/19)^""+6)}}}{{{""=""}}}{{{4(7(22/19)^""-1)}}}

{{{3((66/19)^""+6)}}}{{{""=""}}}{{{4((154/19)^""-1)}}}

{{{3((66/19)^""+(6/1))}}}{{{""=""}}}{{{4((154/19)^""-(1/1))}}}

{{{3((66/19)^""+((6*19)/(1*19)))}}}{{{""=""}}}{{{4((154/19)^""-((1*19)/(1*19)))}}}

{{{3((66/19)^""+(114/19))}}}{{{""=""}}}{{{4((154/19)^""-((19)/(19)))}}}

{{{3((66+114)/19))}}}{{{""=""}}}{{{4((154-19)/(19)))}}}

{{{3((180)/19))}}}{{{""=""}}}{{{4((135)/(19)))}}}

{{{540/19}}}{{{""=""}}}{{{540/19}}}

Trouble is, when you get a terribly ugly solution such as {{{22/19}}},
to check the terribly ugly answer is much harder than solving
the equation.

Edwin</pre>