Algebra 2: solving 0.33x + 0.02y = -1, 0.02x + 0.05y = 14

[(onfus3d]

you have to solve this system using substitution or elimination....

0.33x+0.02y=-1
0.02x+0.05y=14

which method would be best to use?

 

[tkhunny]

The VERY BEST is the one you understand.

 

[soroban]

Re: Algebra 2: solving 0.33x + 0.02y = -1, 0.02x + 0.05y = 1

Hello, (onfus3d!

Solve this system using substitution or elimination.

. . . \(\displaystyle \begin{array}{cc}0.33x\,+\,0.02y &= &-1 \\
0.02x\,+\,0.05y& =& 14\end{array}\)

Substitution often involves fractions . . . I would use elimination.

Multiply both equations by 100: \(\displaystyle \:\begin{array}{cc}33x\,+\,2y & = & -100 \\ 2x\,+\,5y & = & 1400\end{array}\;\;\begin{array}{cc}[1]\\[2]\end{array}\)

Multiply [1] by \(\displaystyle 5:\;\;\;165x\,+\,10y \;=\;-500\)
Multiply [2] by -\(\displaystyle 2:\;\;-4x\,-\,10y\;=\;-2800\)

Add: \(\displaystyle \,161x \:=\:-3300\;\;\Rightarrow\;\;\fbox{x\:=\:-\frac{3300}{161}}\)

Substitute into [2]: \(\displaystyle \:2\left(-\frac{3300}{161}\right)\,+\,5y\:=\:-1400\;\;\Rightarrow\;\;\fbox{y \:=\:\frac{46,400}{161}}\)