Question

NO LINKS!!!
Help me with this problem​

Answer 1

Answer: b = 2.5, c = -2.5

Equation's:

1. 3b - 45c = 120
2. b - 3c = 10

Make b the subject:

3b - 45c = 120

3b = 120 + 45c

 b = (120+ 45c)/3

    = 40 + 15c          ____ equation 1

$\rule{100}{1}$

b - 3c = 10

b = 10 + 3c           ____ equation 2

Solve them Simultaneously:

        b = b

10 + 3c = 40 + 15c

3c - 15c = 40 - 10

-12c = 30

c = -2.5

For b: 10 + 3c = 10 + 3(-2.5) = 2.5

Answer 2

Answer:

c)  b = 2.5, c = -2.5

Step-by-step explanation:

Given system of equations:

$\begin{cases}3b-45c=120\\b-3c=10\end{cases}$

Rearrange the second equation to make b the subject:

$\implies b=10+3c$

Substitute this into the first equation and solve for c:

$\implies 3(10+3c)-45c=120$

$\implies 30+9c-45c=120$

$\implies 30-36c=120$

$\implies -36c=90$

$\implies c=-2.5$

Substitute the found value of c into the rearranged second equation and solve for b:

$\implies b=10+3(-2.5)$

$\implies b=10-7.5$

$\implies b=2.5$

Therefore, the solution to the system of equations is:

$b = 2.5, \:\:\:c = -2.5$