Start with the equation with the fewest variables, \(\displaystyle \ 6z=-12\).

Solve for \(\displaystyle z\) by dividing both sides of the equaion by 6:

\(\displaystyle \frac{6z}{6}=\frac{-12}{6}\)

\(\displaystyle z=-2\)

Plug this \(\displaystyle z\) value into the second equation to solve for \(\displaystyle y\):

\(\displaystyle 9y-5z=-8\)

\(\displaystyle 9y-5(-2)=-8\)

\(\displaystyle 9y+10=-8\)

Subtract 10 from both sides:

\(\displaystyle 9y+10-10=-8-10\)

\(\displaystyle 9y=-18\)

Divide by 9:

\(\displaystyle \frac{9y}{9}=\frac{-18}{9}\)

\(\displaystyle y=-2\)

Plug these \(\displaystyle y\) and \(\displaystyle z\) values into the first equation to find \(\displaystyle x\):

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

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

Combine like terms:

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

Subtract 2:

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

\(\displaystyle -4x=5\)

Divide by -4:

\(\displaystyle \frac{-4x}{-4}=\frac{5}{-4}\)

\(\displaystyle x=-\frac{5}{4}\)

Therefore the final solution is \(\displaystyle (x,y,z)=\left (-\frac{5}{4},-2,-2 \right)\).

\(\displaystyle x\) and \(\displaystyle y\) add up to 16: \(\displaystyle x+y=16\)

When \(\displaystyle x\) is doubled to \(\displaystyle 2x\), they sum to 34: \(\displaystyle 2x+y=34\)

We have two equations and two unknowns, so we can find a solution to this system.

Solve for \(\displaystyle x\) in the first equation:

\(\displaystyle x=16-y\)

Plug this into the second equation:

\(\displaystyle 2(16-y)+y=34\)

Solve for \(\displaystyle y\):

\(\displaystyle 32-2y+y=34\)

\(\displaystyle 32-y=34\)

\(\displaystyle -y=2\)

\(\displaystyle y=-2\)

Use this \(\displaystyle y\) value to find \(\displaystyle x\). We already have a very simple equation for \(\displaystyle x\), \(\displaystyle x=16-y\).

\(\displaystyle x=16-(-2)\)

\(\displaystyle x=18\)

Therefore the answer is \(\displaystyle (x,y)=(18,-2)\).

\(\displaystyle 4x+3y=6\)

\(\displaystyle 2x+2y=4\)

For the second equation, solve for \(\displaystyle x\) in terms of \(\displaystyle y\).

\(\displaystyle 4-2x=2y\)

\(\displaystyle y=2-x\)

Plug this value of y into the first equation.

\(\displaystyle 4x+3(2-x)=6\)

\(\displaystyle 4x + 6- 3x =6\)

\(\displaystyle x=0\)

\(\displaystyle x+y=32\) rearranges to

\(\displaystyle x=32-y\)

and

\(\displaystyle x=\frac{2}{3}y-2\), so

\(\displaystyle 32-y=\frac{2}{3}y-2\)

\(\displaystyle 32-1\frac{2}{3}y=-2\)

\(\displaystyle -1\frac{2}{3}y=-34\)

\(\displaystyle -\frac{5}{3}y=-34\)

\(\displaystyle y=\frac{-34}{1}\cdot\frac{-3}{5}=\frac{102}{5}=20\frac{2}{5}\)

\(\displaystyle x=32-20\frac{2}{5}=11\frac{3}{5}\)

In the second equation, you can substitute \(\displaystyle 3x + 4\) for \(\displaystyle y\) from the first.

\(\displaystyle 2x + 3y = 34\)

\(\displaystyle 2x + 3 (3x + 4) = 34\)

\(\displaystyle 2x + 3 (3x) + 3 (4) = 34\)

\(\displaystyle 2x + 9x + 12 = 34\)

\(\displaystyle 11x + 12 = 34\)

\(\displaystyle 11x = 22\)

\(\displaystyle x = 2\)

Now, substitute 2 for \(\displaystyle x\) in the first equation:

\(\displaystyle y = 3x + 4\)

\(\displaystyle y = 3 (2) + 4\)

\(\displaystyle y = 6 + 4\) 

\(\displaystyle y = 10\)

The solution is \(\displaystyle (2, 10)\)

To find the point where these two lines intersect, set the equations equal to each other, such that \(\displaystyle y\) is substituted with the \(\displaystyle x\) side of the second equation. Solving this new equation for \(\displaystyle x\) will give the \(\displaystyle x\)-coordinate of the point of intersection.

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

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

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

\(\displaystyle 2x = - 2\)

Divide both sides by 2.

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

\(\displaystyle x = - 1\)

Now substitute \(\displaystyle -1\) into either equation to find the \(\displaystyle y\)-coordinate of the point of intersection.

\(\displaystyle y = 3x\)

\(\displaystyle y= 3(-1)\)

\(\displaystyle y = -3\)

With both coordinates, we know the point of intersection is \(\displaystyle (-1,-3)\). One can plug in \(\displaystyle -1\) for \(\displaystyle x\) and \(\displaystyle -3\) for \(\displaystyle y\) in both equations to verify that this is correct.

\(\displaystyle 3x - 5y = 5\)

\(\displaystyle -2x + 5y = 0\)

Add the equations together.

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

\(\displaystyle -5y+5y=0\)

\(\displaystyle 5+0=5\)

Put the terms together to see that \(\displaystyle x+0=5\ \textup{or}\ x=5\).

Substitute this value into one of the original equaitons and solve for \(\displaystyle y\).

\(\displaystyle 3x - 5y = 5\)

\(\displaystyle 3(5) - 5y = 5\)

\(\displaystyle -5y=-10\)

\(\displaystyle y=2\)

Now we know that \(\displaystyle x=5\ \textup{and}\ y=2\), thus we can find the sum of \(\displaystyle x\) and \(\displaystyle y\).

\(\displaystyle x+y=5+2=7\)

We can solve this problem by setting up a simple system of equations. First, we want to change the equations so one variable can cancel out. Multiplying the first equation by 2 and the second equation by 3 gives us a new system of \(\displaystyle 2y = 6x + 8\) and \(\displaystyle 3y=-6x+27\). These equations add up to \(\displaystyle 5y=35\) or \(\displaystyle y = 7\). Plugging in 7 for \(\displaystyle y\) in either of the original two equations shows us that \(\displaystyle x\) is equal to 1 and the point is \(\displaystyle (1,7)\).

This system has infinite solutions becasue the two equations are actually the exact same line. To discover this, put both equations in terms of y.

First, \(\displaystyle 2x - y = -3\). Add y to both sides:

\(\displaystyle 2x = -3 + y\) Now add 3 to both sides:

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

Now we can show that the second equation also represents the line \(\displaystyle y=2x + 3\)

\(\displaystyle 2y-6 = 4x\) add 6 to both sides

\(\displaystyle 2y = 4x + 6\) divide both sides by 2

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

Since both equations are the same line, literally any point on one line will also be on the other - infinite solutions.

A system of equations can be solved by subsituting one variable for another. Since we know that \(\displaystyle y=x-2\), we can subsitute this into the other equation so \(\displaystyle x-2=2x-4\). This expression can be solved to find that \(\displaystyle x=-6\). Now that we know the value of \(\displaystyle x\) it can be subsituted into either of the original equations to find \(\displaystyle y\).