To solve this equation you will plug the second equation straight into the first one by substituting what is written there for $\displaystyle y$.

You will then get:

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

From here you need to simplify by combining like terms:

$\displaystyle 4x-2=2$

Bring the $\displaystyle -2$ over by addition:

$\displaystyle 4x=4$

Then divide both sides by $\displaystyle 4$ to get:

$\displaystyle x=1$

You will then take this value of $\displaystyle x$ and plug it into either equation.

$\displaystyle y=(1)-2=-1$

Start by making a list of the multiples of 4 that are smaller than the square of 7. When 7 is squared, it equals 49; thus, we can compose the following list:

$\displaystyle \text{4, 8, 12, 16, 20, 24, 28, 32, 36, 40, 44, and 48}$

Next, make a list of all the multiples of 6 that are greater than 0. Since we are looking for shared multiples, stop after 48 because numbers greater than 48 will not be included in set A. The biggest multiple of 4 smaller that is less than 49 is 48; therefore, do not calculate multiples of 6 greater than 48.

$\displaystyle \text{6, 12, 18, 24, 30, 36, 42, and 48}$

Finally, count the number of multiples found in both sets. Both sets include the following numbers:

$\displaystyle \text{12, 24, 36, and 48}$

The correct answer is 4 numbers.

Since the object in question is a cube, each of its sides must be the same length. Therefore, to get a volume of $\displaystyle 216$ $\displaystyle cm^{3}$, each side must be equal to the cube root of $\displaystyle 216$, which is $\displaystyle 6$ cm.

We can then set each expression equal to $\displaystyle 6$.

The first expression $\displaystyle (3x^2^$ $\displaystyle -6)$ can be solved by either $\displaystyle -2$ or $\displaystyle 2$, but the other two expressions make it evident that the solution is $\displaystyle x = -2$.

The most simple method for solving systems of equations is to transform one of the equations so it allows for the canceling out of a variable. In this case, we can multiply $\displaystyle 3x + y = 8$ by $\displaystyle (-4)$ to get $\displaystyle -12x - 4y = -32$.

 Then, we can add $\displaystyle 2x + 4y = 12$ to this equation to yield $\displaystyle -10x = -20$, so $\displaystyle x = 2$.

We can plug that value into either of the original equations; for example, $\displaystyle 3(2) )+ y = 8$.

So, $\displaystyle y = 2$ as well.

By solving one equation for $\displaystyle y$, and replacing $\displaystyle y$ in the other equation with that expression, you generate an equation of only 1 variable which can be readily solved.

Multiply the bottom equation by 5, then add to the top equation:

$\displaystyle 6x-y = 24$

$\displaystyle 5 \left (6x-y \right ) =5\cdot 24$

$\displaystyle 30x-5y =120$

$\displaystyle \underline{\textrm{\; } 3x + 5y = \; \; 1}$

$\displaystyle 33x \;\;\;\;\;\; \; =121$

$\displaystyle x = \frac{121}{33} = \frac{11}{3} = 3 \frac{2}{3}$

Multiply the top equation by $\displaystyle -2$:

$\displaystyle 3x + 5y = 23$

$\displaystyle -2 \cdot \left (3x + 5y \right )= -2 \cdot 23$

$\displaystyle -6x -10y \right )= -46$

Now add:

   $\displaystyle 6x-\; \; y = -9$

$\displaystyle \underline{-6x -10y = -46}$

$\displaystyle -11y = - 55$

$\displaystyle -11y \div (-11) = -55\div (-11)$

$\displaystyle y = 5$

Multiply the top equation by $\displaystyle -2$:

$\displaystyle 3x + 5y = 1$

$\displaystyle -2 \cdot \left (3x + 5y \right )= -2 \cdot 1$

$\displaystyle -6x -10y \right )= -2$

Now add:

    $\displaystyle \; \; 6x-\; \; \; y = 24$

$\displaystyle \underline{-6x -10y= -2} \right )$

          $\displaystyle -11y = 22$

$\displaystyle -11y \div (-11) = 22\div (-11)$

$\displaystyle y = -2$

$\displaystyle a + 3b = 5$

$\displaystyle a-2b = 0$

To solve this system of equations, use substitution. First, convert the second equation to isolate $\displaystyle \small a$.

$\displaystyle a-2b = 0\rightarrow a=2b$

Then, substitute $\displaystyle \small 2b$ into the first equation for $\displaystyle \small a$.

$\displaystyle a + 3b = 5$

$\displaystyle (2b) + 3b = 5$

Combine terms and solve for $\displaystyle \small b$.

$\displaystyle 5b=5$

$\displaystyle b=1$

Now that we know the value of $\displaystyle \small b$, we can solve for $\displaystyle \small a$ using our previous substitution equation.

$\displaystyle a=2b=2(1)=2$

When we add the two equations, the $\displaystyle x$ and $\displaystyle y$ variables cancel leaving us with:

$\displaystyle 1=8$   which means there is no solution for this system.