x = 5; y = 12; z = a + b;

x

z

x y^2

Expand[z^x]

Clear[x,y,z]

x

f=x^2+3x+2; g=(2x+3)(5x-2);h=2/(x-3)-1/(x-1);k=(x+5)/((x+2)(x+3));

f

Factor[f]

Expand[g]

Together[h]

Apart[k]

Clear[f,g,h,k]

Solve[3x-8==4]

Solve[z^2+2z==-2]

Solve[a t-2==3a, t]

Simplify[%]

f = x^2-3x-4; Solve[f==0]

Clear[f]

g = 1/(x-2); h = 4/(x+1);

Solve[g==h]

Clear[g,h]

Solve[{x-2y==4,x-1==5y}, {x,y}]

Solve[{3x-4y==4,x+2y==7}, {x,y}]

f = 2x+y; g = 5x+2y;

Solve[{f==2, g==3},{x,y}]

Clear[f,g]

NSolve[2x-3==x^2-3x-4]

NSolve[2x-3==x^2-3x-4, x]

NSolve[{x^2+y^2==16, y==x^2-2x+2}, {x,y}]

f[x_]:=x^2 - 5; g[x_]:=Log[x]/x

f[3]

g[f[x]]

g[f[3]]

N[%]

Solve[f[x]==4]

Clear[f,g]

f[x_]:=1/(x-1); g[x_]:=(x+1)/x

f[g[x]]

Simplify[%]

g[f[x]]

Simplify[%]

Functions  f and g are inverses of one another!

Clear[f,g]

Plot[x^2-3x+4,{x,-10,10}]

Plot[Log[Abs[x]],{x,-6,6}]

y = x^2-3x-10; Plot[y,{x,-5,8}]

Clear[y]

f[x_]:=x+E^(-x); Plot[f[x], {x,-4,6}]

The  y-coordinate  of the point with the x-coordinate 1 is approximately 1.46.

f[1]//N

Clear[f]

f[x_]:=-x^2+4x+5; g[x_]:=E^(x)

Plot[{f[x],g[x]},{x,-2,4}]

The  points of intersection are approximately (-0.9, 0.5) and 
(2.2, 9.2).

NSolve[f[x]==g[x]]

Clear[f,g]