Solve the system: { 0.01 − x 0.02 − 0.5 ≤ 2 − 3 x 0.01 3 − x 0.1 ≤ 6 x + 8 0.4 − 5 Is it a good idea to multiply by 10?

Question

Solve the system:      {                                                                      0.01                −                x                            0.02                                −          0.5          ≤                                                    2                −                3                x                            0.01                                                                                                      3                −                x                            0.1                                ≤                                                    6                x                +                8                            0.4                                −          5                        Is it a good idea to multiply by 10?

Zachary Washington · Accepted Answer

Step 1I would multiply the first fraction by 100 on top and bottom to get            1      −      100      x        2    .The second fraction is       1    2   and if you multiply the third fraction by 100 top and bottom you get   200  −  300  x . So the first inequality becomes            1      −      100      x        2    +      1    2    ≤  200  −  300  x  .Multiply through by 2 to finish clearly fractions to get  1  −  100  x  +  1  ≤  400  −  600  x  500  x  ≤  398  x  ≤  0.796.Treat the second inequality the same way, multiplying each fraction, top and bottom by 10.

gravplasszo · Accepted Answer

Step 1If you multiply the first equation&#039;s denominator by 50, you get:  (  0.01  −  x  )  −  0.001  ≤  4  −  6  xRemember that each zero in a decimal is a power of ten and the rest you can see as numbers from   1  −  9 . Like what I did with the first fraction on the LHS, I multiplied the denominator by 10 first which gave me             0.01      −      x        0.2   and all I did was then multiply by 5 which gave me   0.01  −  x  .Remember that you have to multiply the rest of the equation by that amount too - that&#039;s why the RHS multiplied by 2 since   0.01  ∗  50  =  0.5 which means   0.01  −  x so I have to multiply the numerator by 2.See if you can do the second equation using this method - remember to look for an amount which will eliminate all the fractions. I&#039;m thinking 10 for the second equation, but I&#039;ll let you take it from there.

Solve the system: {((0.01-x)/(0.02)-0.5le(2-3x)/(0.01)),((3-x)/(0.1)le(6x_8)/(0.4)-5):} Is it a good idea to multiply by 10?

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