Question

Math algebra two plz show your work

Answer 1

Answer:

The answer is $b=3, a=-2$, and $c=3$.

Step-by-step explanation:

To solve this system of equations, start by solving for (a) in the third equation.

To solve for (a) in the third equation, add $3b$ to both sides of the equation, which will look like $2a=-13+3b\\-a+b-c=2\\2a+3b-4c=-7$. Next, divide each term in $2a=-13+3b$ by 2 and simplify, which will look like $\frac{2a}{2}=\frac{-13}{2} +\frac{3b}{2} \\-a+b-c=2\\2a+3b-4c=-7$   =  $a=\frac{-13}{2} +\frac{3b}{2} \\-a+b-c=2\\2a+3b-4c=-7$.

Then, replace all variables of (a) with $-\frac{13}{2} +\frac{3b}{2}$ in each equation and simplify, which will look like $-13+6b-4c=-7\\-\frac{2c-13+b}{2}=2\\a=-\frac{13}{2}+\frac{3b}{2}$.

The next step is to reorder $-\frac{13}{2}$ and $\frac{3b}{2}$, which will look like $\frac{3b}{2}-\frac{13}{2}\\-13+6b-4c=-7\\-\frac{2c-13+b}{2} =2$.

Then, solve for (b) in the second equation. To solve for (b) in the second equation start by moving all terms not containing (b) to the right side of the equation, which will look like $6b=6+4c\\a=\frac{3b}{2}-\frac{13}{2} \\-\frac{2c-13+b}{2} =2$. Next, divide each term in                ($6b=6+4c$) and simplify, which will look like $b=1+\frac{2c}{3} \\a=\frac{3b}{2} -\frac{13}{2\\}\\-\frac{2c-13+b}{2} =2$.

Then, replace all variables of (b) with $1+\frac{2c}{3}$ in each equation and simplify, which will look like $-\frac{2(2c-9)}{3}=2\\a=c-5\\b=1+\frac{2c}{3}$.

The next step is to solve for (c) in the first equation. To solve for (c) in the first equation start by multiplying both sides of the equation by $-\frac{3}{2}$ and simplify, which will look like $2c-9=-3\\a=c-5\\b=1+\frac{2c}{3}$. Then, move all terms not containing (c) to the right side of the equation, which will look like $2c=6\\a=c-5\\b=1+\frac{2c}{3}$. Next, divide each term in $2c=6$ by 2 and simplify, which will look like $c=3\\a=c-5\\b=1+\frac{2c}{3}$.

Then, replace all variables of (c) with 3 in each equation and simplify, which will look like $b=3\\a=-2\\c=3$. Finally, the list of all the solutions are $b=3,a=-2$, and $c=3$.