<h3>6(x + 1)-5x = 8 + 2(x - 1)</h3>
<em>We move all terms to the left:</em>
<h3>6(x + 1) -5x -(8 + 2(x - 1)) = 0</h3>
<em>We add all the numbers and all the variables.</em>
<h3>-5x + 6(x + 1) -(8 + 2(x - 1)) = 0</h3>
<em>Multiply</em>
<h3>-5x + 6x -(8 + 2(x - 1)) + 6 = 0</h3>
<em>Calculations in parentheses: </em><em>-(8 + 2(x - 1))</em><em>, so:</em>
<h3>8 + 2(x - 1)</h3>
<em>DeterminingTheFunctionDomain</em>
<h3>2(x - 1) + 8</h3>
<em>Multiply</em>
<h3>2x - 2 + 8</h3>
<em>We add all the numbers and all the variables.</em>
<h3>2x + 6</h3>
<em>Back to the equation</em>:
<h3>-(2x + 6)</h3>
<em>We add all the numbers and all the variables.</em>
<h3>x-(2x + 6) + 6 = 0</h3>
<em>We get rid of the parentheses.</em>
<h3>x - 2x - 6 + 6 = 0</h3>
<em>We add all the numbers and all the variables.</em>
<h3>-x = 0</h3><h3>x = 0/-1</h3><h3>x = 0</h3>
<em>Therefore, </em><em>this exercise</em><em> has </em><em>no real solution.</em>