This problem addresses some common algebraic errors. for the equalities stated below assume that x and y stand for real numbers. assume that any denominators are non-zero. mark the equalities with t (true) if they are true for all values of x and y, and f (false) otherwise.(x+y)^2 = x^2 + y^2.(x+y)^2 = x^2 + 2xy+y^2.x/ (x+y)= 1/y. x-(x+y) = y. sqrt{x^2} = x. sqrt{x^2} = |x|.sqrt{x^2+4} = x+2.1(x+y) = 1/x + 1/y.

The answers are:

a) is FALSE.

b) is TRUE.

c) is FALSE.

d) is FALSE.

e) is FALSE.

f) is TRUE.

g) is FALSE.

h) is FALSE.

Step-by-step explanation:

We can show that the FALSE statement are, in fact, false finding an example where the equality fails.

a) Take x=2 and y=3 and notice that (2+3)² = 25, while 2²+3²=4+9=13.

b) (x+y)² = (x+y)(x+y)=x²+xy+xy+y² = x² +2xy +y².

c) Take x=2 and y=3 and notice that 2/(2+3) = 2/5 which is different from 1/3.

d) Take x=2 and y=3 and notice that 2-(2+3) = 2-5=-3 which is different from 3. Recall that x-(x+y) = x-x-y=-y.

e) Take x=-2, and notice that [tax]\sqrt{(-2)^2} = \sqrt{4} = 2[/tex] which is different from -2.

f) The previous example illustrate why this is true.

g) Take x=1, and notice that which is different from 3.

h) Take x=2 and y=3 and notice that 1/(2+3)=1/5 and 1/3+1/2 = 5/6.

are ya tryin to ask something or is this a statement

answer:

given the linear inequality :

add 2y to both sides we get;

simplify:

add 12 to both sides of an equation;

simplify:

divide both sides by 2 we get;

or

the related equation for the inequality equation [1] is;

y-intercepts defined as the graph crosses the y-axis.

substitute x =0 in [2] to solve for y;

y = 6

x-intercepts defined as the graph crosses the x-axis.

substitute value y = 0 to solve for x;

subtract 6 from both sides we get

multiply both sides by 2 we get;

or

x = -12

now, plot these point (0, 6) and (-12, 0) on graph as shown below and shade the graph below the solid line.