\(\displaystyle 4 (x +6) + 11 = 6 (x + 7) - x\)

\(\displaystyle 4 \cdot x +4 \cdot 6 + 11 = 6 \cdot x + 6 \cdot 7 - x\)

\(\displaystyle 4 x +24 + 11 = 6x + 42 - x\)

\(\displaystyle 4 x +35 = 5x + 42\)

\(\displaystyle 4 x +35 -4x -42 = 5x + 42-4x -42\)

\(\displaystyle -7 = x\)

\(\displaystyle x^2-7x+3=x^2-6x-2\)

Notice that \(\displaystyle x^2\) can be canceled out by subtracting it from both sides.

\(\displaystyle x^2-x^2-7x+3=x^2-x^2-6x-2\)

\(\displaystyle -7x+3=-6x-2\)

Now, we can isolate \(\displaystyle x\). First, we need to get all the variables on the same side by adding \(\displaystyle 6x\) to each side.

\(\displaystyle -7x+6x+3=-6x+6x-2\)

\(\displaystyle -x+3=-2\)

Subtract \(\displaystyle 3\) from both sides.

\(\displaystyle -x+3-3=-2-3\)

\(\displaystyle -x=-5\)

Finally, multiply both sides by \(\displaystyle -1\).

\(\displaystyle -1(-x)=-1(-5)\)

\(\displaystyle x=5\)

\(\displaystyle 5 (x - 4) + 5 (x - 5) = 2 (x - 4) + 4 (2x - 7)\)

\(\displaystyle 5 \cdot x - 5 \cdot 4 + 5 \cdot x - 5 \cdot 5 = 2 \cdot x - 2 \cdot 4 + 4 \cdot 2x - 4 \cdot 7\)

\(\displaystyle 5x - 20 + 5 x - 25 = 2 x - 8 + 8x - 28\)

\(\displaystyle 10x - 45 = 10x - 36\)

\(\displaystyle 10x - 45 - 10x = 10x - 36- 10x\)

\(\displaystyle - 45 = - 36\)

This statement is identically false, so the original equation has no solution.

\(\displaystyle \frac{3}{4} x = \frac{5}{12} x + \frac{2}{3}\)

\(\displaystyle 12 \cdot \frac{3}{4} x =12 \cdot \left ( \frac{5}{12} x + \frac{2}{3} \right )\)

\(\displaystyle \frac{12}{1} \cdot \frac{3}{4} x =\frac{12}{1}\cdot \frac{5}{12} x +\frac{12}{1}\cdot \frac{2}{3}\)

\(\displaystyle \frac{3}{1} \cdot \frac{3}{1} x =\frac{1}{1}\cdot \frac{5}{1} x +\frac{4}{1}\cdot \frac{2}{1}\)

\(\displaystyle 9x =5x +8\)

\(\displaystyle 9x-5x =5x -5x +8\)

\(\displaystyle 4x =8\)

\(\displaystyle 4x \div 4=8\div 4\)

\(\displaystyle x = 2\)

\(\displaystyle \frac{3}{5} x + \frac{3}{20} = \frac{4}{5}\)

\(\displaystyle 20 \cdot \left (\frac{3}{5} x + \frac{3}{20} \right ) =20 \cdot \frac{4}{5}\)

\(\displaystyle \frac{20}{1} \cdot \frac{3}{5} x + \frac{20}{1} \cdot \frac{3}{20} =\frac{20}{1} \cdot \frac{4}{5}\)

\(\displaystyle \frac{4}{1} \cdot \frac{3}{1} x + \frac{1}{1} \cdot \frac{3}{1} =\frac{4}{1} \cdot \frac{4}{1}\)

\(\displaystyle 12x+3=16\)

\(\displaystyle 12x+3-3=16-3\)

\(\displaystyle 12x=13\)

\(\displaystyle 12x\div 12=13\div 12\)

\(\displaystyle x = \frac{13}{12}\)

\(\displaystyle \frac{1}{2} x - 4 = \frac{3}{10}x + \frac{2}{5}\)

\(\displaystyle 10 \cdot \left (\frac{1}{2} x - 4 \right )=10 \cdot \left ( \frac{3}{10}x + \frac{2}{ 5} \right )\)

\(\displaystyle \frac{10}{1} \cdot \frac{1}{2} x - \frac{10}{1} \cdot \frac{4}{1} \right )=\frac{10}{1} \cdot \frac{3}{10}x +\frac{10}{1} \cdot \frac{2}{ 5}\)

\(\displaystyle \frac{5}{1} \cdot \frac{1}{1} x - \frac{10}{1} \cdot \frac{4}{1} \right )=\frac{1}{1} \cdot \frac{3}{1}x +\frac{2}{1} \cdot \frac{2}{ 1}\)

\(\displaystyle 5x-40 = 3x+4\)

\(\displaystyle 5x-40 -3x +40= 3x +4 -3x +40\)

\(\displaystyle 2x=44\)

\(\displaystyle x=22\)

\(\displaystyle 5 (x - 4) + 2 (x - 3) = 3 (x - 4) + 2 (2x - 7)\)

\(\displaystyle 5 \cdot x -5 \cdot 4 + 2 \cdot x - 2 \cdot 3 = 3 \cdot x - 3 \cdot 4 + 2 \cdot 2x - 2 \cdot 7\)

\(\displaystyle 5x - 20 + 2 x - 6 = 3x - 12 + 4x - 14\)

\(\displaystyle 7x - 26 = 7x - 26\)

This statement is identically true; therefore, the solution set is the set of all real numbers.

Call the measure of \(\displaystyle \angle1\) \(\displaystyle x\). Since the measure of \(\displaystyle \angle2\) is twenty degrees more than that of \(\displaystyle \angle 1\), we can say that \(\displaystyle \angle2\) has measure \(\displaystyle x + 2 0\). Since the measure of \(\displaystyle \angle3\) is forty degrees less than twice that of \(\displaystyle \angle 1\), we can say that \(\displaystyle \angle3\) has measure \(\displaystyle 2x-40\). 

The sum of the measures of the three angles of a triangle is 180, so we can set up and solve an equation that sets the sum of these three expressions equal to 180:

\(\displaystyle x + (x + 20) + (2x - 40) = 180\)

\(\displaystyle x + x+ 2x + 20 - 40= 180\)

\(\displaystyle 4x - 20 = 180\)

\(\displaystyle 4x = 200\)

\(\displaystyle x = 50\)

Substitute 50 for \(\displaystyle x\) in the other two expressions:

\(\displaystyle x + 20 = 50 + 20 = 70\)

\(\displaystyle 2x-40 = 2\cdot50 - 40 = 60\)

The triangle has angles measuring \(\displaystyle 50^{\circ }, 60^{\circ },70^{\circ }\), so the answer to the question is \(\displaystyle 70^{\circ }\).

\(\displaystyle 5h + 7k = 12\)

\(\displaystyle 5h + 7k - 7k = 12- 7k\)

\(\displaystyle 5h = 12- 7k\)

\(\displaystyle \frac{5h}{5} =\frac{ 12- 7k}{5}\)

\(\displaystyle h =\frac{ - 7k+12}{5}\)

\(\displaystyle 2x+8=13\)

Solve equation by using inverse order of operations to get \(\displaystyle x\) by itself.

Order of operations: PEMDAS. Addition and subtraction usually come last, so when using inverse order of operations to "undo an expression," they come first. 8 is being added to \(\displaystyle x\), so subtract 8 from both sides in order to get \(\displaystyle x\) alone.

\(\displaystyle 2x+8=13\)

      \(\displaystyle -8\)

\(\displaystyle 2x=5\)

Next comes multiplication/division. \(\displaystyle x\) is being multiplied by 2, so divide both sides by 2 to get \(\displaystyle x\) on its own.

\(\displaystyle \frac{2x}{2}=\frac{5}{2}\)

\(\displaystyle x=\frac{5}{2}\)