Paul,
Welcome to Maple and good luck with your efforts!
Here's one way to do what you want. Copy and paste the following code into a new worksheet and click !!! on the Toolbar:
restart;
x := 3/2*y - 1/2;
'x' = seq(evalf ( eval(x,y=i)),i in [-2,0,3/2,3]);
If you would like to show the answers expressed as rational numbers, just delete "evalf" from the last line of code.
I don't think there is a step by step way of showing this, but have a look at Tools>Tutors>Pre-Calculus and see if there is something there that you think might be suitable
Hope this helps.
J. Tarr

John,
You could also try something like this;
restart;
X:=[380,4000,4067,5789];
Y:=[10, 35, 56,82];
plots[pointplot](zip((x,y) ->[x,y],X,Y),symbol=diagonalcross);
Hope this helps.
J. Tarr

You could do something like this, where L is your list of values:
restart;
min(seq(L[i],i=1..nops(L)));
Hope this helps.
J. Tarr

Do you think it is courteous to demand an answer to your problem a mere 50 minutes after first posting it? Perhaps you should learn patience in addition to Maple.
Please see a reply to your other post for the answer to your problem.
J. Tarr

If you are dealing with a problem in control engineering, I would suggest that you list all your assumptions and linearizations before you try to write any equations for the systems. Then use these assumptions/linearizations to write your systems' equations, which should keep them as simple (and as soluble) as possible. If the solutions don't seem right, you can always go back to the assumptions and linearizations and add some more realism at the expense of complication and, may be, solubility.
J. Tarr

Robert,
Could you please post again?
Many thanks.
J. Tarr

Hendrick,
You could do something like this, but really you don't need Maple for it:
restart;
eq1 := 2400*a + 1800*b = 81000;
eq2 := a + b = 40;
solution1 := solve({eq1,eq2},{a,b});
eq3 := 525*c + 600*d = 7500;
eq4 := c + d = 13;
solution2 := solve({eq3,eq4},{c,d});
Hope this helps.
J. Tarr

Could you please upload your worksheet? There seems to be something wrong with the extract posted above: G11 is zero because psi2 is not a function of a[1], and similarly for dpsi2.
J. Tarr

Florian asked why assigning the value 6 to the variable p *before* solving the equation for F yielded two real solutions, whereas assigning that value to p *after* solving for F, produced 3 complex solutions.
True, the second method produces imaginary parts of F that are all very small, and two of the real parts by both methods are identical. But why does the first method produce only two real solutions?
J. Tarr

Try changing subticks from 10 to 8 (the non-negative integer specifying the exact number of subticks between two gridlines or tickmarks). Please see ?plot,tickmarks
Hope this helps.
J. Tarr

It may be helpful to remember that the term “tensor” first arose in the study of elasticity. Indeed, it was the physicists, J.W. Gibbs and J.C. Maxwell, who mainly initiated tensor theory. The late Professor A.P. Wills of Columbia University gave an excellent history of tensor theory in his book

*Vector Analysis with an Introduction to Tensor Analysis* (1931).
This clue leads us to the

Stress Tensor in Wikepedia. IMHO this gives a very reasonable visualization of the stress tensor and, by extension (no pun intended!), other tensors.
Hope this helps.
J. Tarr

Jean,
Another way of achieving only 3 decimal places is to use interface(displayprecision=3). See the example below.
You mentioned array and matrix. In Maple these are different objects, and although you can continue to use them, the newer versions of these - Array and Matrix - are more efficient. Please see ?Array and ?Matrix.
Here are two examples of Matrices filled with the initial values that you specified. If you want to see the contents of B, right click on the Matrix placeholder in the output and select Browse.
restart;
interface(displayprecision=3):
f := (i,j) -> `if`(j<=2,i^j,evalf(i^(1/j))):
A := Matrix(1..10,1..3,f);
B := Matrix(1..1000,1..3,f);
Hope this helps.
J. Tarr

Statistics[Distributions][LogNormal] might be of some use, e.g.
`CDF(RandomVariable(LogNormal(mu,sigma)),x)` = piecewise(x < 0, 0, x < exp(mu), 1/2-(1/2)*erf((1/2)*(mu-ln(x))*sqrt(2)/sigma), 1/2+(1/2)*erf((1/2)*(ln(x)-mu)*sqrt(2)/sigma))
Hope this helps.
J. Tarr

Could you please clarify what i and j are in your two equations?
J. Tarr

Jean-Marc's suggestion is correct. You could achieve the same thing with:
restart;
plots[logplot](10^n, n = 1..4,tickmarks=[4,40])
It's possible to exercise a great deal of control over the scaling of the axes tickmarks and the labels that appear beside them. Please see ?plot/options, ?plot/tickmarks and ?plot/typesetting.
Hope this helps.
J. Tarr