Question

solve by substitution
r+s=-12
4r-6s=12

Answer 1

Rearrange the first equation by subtracting s from both sides.
notice that you have an r = something?
usbstitute that r into the 2nd equation and solve for s.
once you have ur S, plug that value in for the original equation and solve for r.

Answer 2

okay

Answer 3

wait show me how please?

Answer 4

For example, \[ax + by = c\]
and \[dx + ey = f\]
I would rearrange the first equation to make it only have Y's so i CAN substitute in to the second, to thus have:
\[x = \frac{ c }{ a }-\frac{ by }{ a }\]
now i can substitute that x into the second equation and solve for y: \[d(\frac{ c }{ a }-\frac{ by }{ a })-ey = f\]
I would distribute the d across to everything:
\[\frac{ dc }{ a }-\frac{ dby }{ a }-ey=f = \frac{ dc-dby }{ a}-ey = ...\]
and so forth to solve for Y!
once you have y, you can substitute that value into the first one, which was: \[ax + by = c\]

Answer 5

i guess

Answer 6

I can't give answer if I don't think you're trying. Did you at least try?

Answer 7

i dont want the answer just want to understand it with the same problem not an example?

Answer 8

ill give you a metal just help me out

Answer 9

\[208x+y = 90\] & \[31372x+\frac{ 45 }{ 117 }y=10000\]
Obviously, the first equation is simpler to rearrange than the second, so you rearrange it using algebra!
\[y = 90-208x\]
substitute THAT new function into the second one! It's the same as saying y = 3 or whatever they tell you.
so you get \[31372x+\frac{ 45 }{ 117 }(90-208x)=10000\]
distribute the fraction across to get only x's. Solving for x, you get:
\[31372x+\frac{ 450 }{ 13 }-80x=100\]
\[31292x+\frac{ 450 }{ 13 }=10000\]
\[31292x = 9965.4\]
and divide to solve for x!
\[x = 0.318464292\]
Now, plug that value into the original equation:
\[208(0.318464292)+y = 90\]
\[y = -23.759\]

Answer 10

Sorry y = 23.759
I gave you a much more complicated example to try and point out everything in some form of detail. This way, hopefully, better prepare you for future problems like this because these types of problems are common and come up a lot in more advance math courses such as algebra, college algebra, pre-calculus, calculus, and beyond (if you plan on going to higher math courses, you will see it in differential equations and learn new methods for solving similar problems in even higher than differential equation courses)

Answer 11

thank you for your help