First, combine like terms to get \(\displaystyle 5x+3=x+11\).

Then, subtract 3 and \(\displaystyle x\) from both sides to get \(\displaystyle 4x=8\).

Then, divide both sides by 2 to get a solution of \(\displaystyle x=4\).

To solve this system of equations, the elimination method can be used (the \(\displaystyle y\) terms cross out).

Once you eliminate the \(\displaystyle y's\), you have \(\displaystyle 12x=-12\).

Then isolate for \(\displaystyle x\), and you get \(\displaystyle x=-1\).

Plug \(\displaystyle -1\) into the first equation to solve for \(\displaystyle y\).

\(\displaystyle 3(-1)+5y=-11\)

\(\displaystyle -3+5y=-11\)

\(\displaystyle -3+5y+3=-11+3\)

\(\displaystyle 5y=-8\)

\(\displaystyle \frac{5y}{5}=\frac{-8}{5}\)

\(\displaystyle y=-\frac{8}{5}\)

Rearrange the second equation:

\(\displaystyle 6y -12x = -42 \Rightarrow 12x - 6y = 42\)

This is just a multiple of the first equation (by a factor of 3).  Therefore, the two equations are dependent on each are and are going to have an infinite number of solutions.

Solve the first equation for \(\displaystyle y\):

\(\displaystyle 3x+2y=12\)

\(\displaystyle 2y=12-3x\)

\(\displaystyle y=6-\frac{3}{2}x\)

Substitute into the second equation:

\(\displaystyle 4x-3(6-\frac{3}{2}x)=-1\)

\(\displaystyle 4x-18+\frac{9}{2}x=-1\)

Multiply the entire equation by 2 to eliminate the fraction:

\(\displaystyle 8x-36+9x=-2\)

\(\displaystyle 17x-36=-2\)

\(\displaystyle 17x=34\)

\(\displaystyle x=2\)

Using the value of \(\displaystyle x\), solve for \(\displaystyle y\):

\(\displaystyle 3(2)+2y=12\)

\(\displaystyle 6+2y=12\)

\(\displaystyle 2y=6\)

\(\displaystyle y=3\)

Therefore, the solution is \(\displaystyle \left ( 2,3 \right )\)

To solve this system of equations, we must first eliminate one of the variables. We will begin by eliminating the \(\displaystyle x\) variables by finding the least common multiple of the \(\displaystyle x\) variable's coefficients.  The least common multiple of 3 and 2 is 6, so we will multiply each equation in the system by the corresponding number, like

\(\displaystyle 2(3x+2y=12)\)

\(\displaystyle 3(-2x+3y=5)\).

By using the distributive property, we will end up with

\(\displaystyle 6x+4y=24\)

\(\displaystyle -6x+9y=15\)

Now, add down each column so that you have

\(\displaystyle 0x+13y=39 \rightarrow 13y=39\)

Then you solve for \(\displaystyle y\) and determine that \(\displaystyle y=3\).

But you're not done yet!  To find \(\displaystyle x\), you have to plug your answer for \(\displaystyle y\) back into one of the original equations:

\(\displaystyle 3x+2(3)=12\)

Solve, and you will find that \(\displaystyle x=2\).  

Set the two equations equal to one another:

2x - 2 = 3x + 6

Solve for x:

x = -8

Plug this value of x into either equation to solve for y.  We'll use the top equation, but either will work.

y = 2 * (-8) - 2

y = -18

Solve the second equation for y:

x - 2y = 4

-2y = 4 - x

y = -2 + x/2

Plug this into the first equation:

3x + 2(-2 + x/2) = 8

Solve for x:

3x - 4 + x = 8

4x = 12

x = 3

Plug this into the second equation to get a value for y:

3 - 2y = 4

2y = -1

y = -0.5

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.

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

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

\(\displaystyle x=0\)