Algebra<span><span>Introduction to Algebra
</span>Variables<span>
</span>Expressions<span>
</span>Equations<span>
</span>Solution of an equation<span>
</span>Simplifying equations<span>
</span>Combining like terms<span>
</span>Simplifying with addition and subtraction<span>
</span>Simplifying by multiplication<span>
</span>Simplifying by division<span>
</span>Word problems as equations<span>
</span>Sequences VariablesA variable is a symbol that represents a number. Usually we use letters such as n, t, or x for variables. For example, we might say that s stands for the side-length of a square. We now treat s as if it were a number we could use. The perimeter of the square is given by 4 × s. The area of the square is given by s× s. When working with variables, it can be helpful to use a letter that will remind you of what the variable stands for: let n be the number of people in a movie theater; let t be the time it takes to travel somewhere; let d be the distance from my house to the park. ExpressionsAn expression is a mathematical statement that may use numbers, variables, or both.Example:The following are examples of expressions:2x3 + 72 × y + 52 + 6 × (4 - 2)z + 3 × (8 - z<span>)</span></span>
Answer:
d = -1/3, 0
Step-by-step explanation:
Subtract the constant on the left, take the square root, and solve from there.
(6d +1)^2 + 12 = 13 . . . . given
(6d +1)^2 = 1 . . . . . . . . . .subtract 12
6d +1 = ±√1 . . . . . . . . . . take the square root
6d = -1 ±1 . . . . . . . . . . . .subtract 1
d = (-1 ±1)/6 . . . . . . . . . . divide by 6
d = -1/3, 0
_____
Using a graphing calculator, it is often convenient to write the function so the solutions are at x-intercepts. Here, we can do that by subtracting 13 from both sides:
f(x) = (6x+1)^ +12 -13
We want to solve this for f(x)=0. The solutions are -1/3 and 0, as above.
70
make 30 minutes into an hour then divide 140 by 2
140/2=70
70x1=70
Question
x+5/x+2 - x+1/x²+2x
Answer:
= (x² - 4x - 1)/[x (x+2)]
= (x² - 4x - 1)/[x² + 2x]
Step-by-step explanation:
x + 5/x + 2 - x + 1/x² + 2x
We factorise the second denominator to give us :
x + 5/x + 2 - x + 1/x(x + 2)
We find the L.C.M of both denominators which is x(x+2).
[x(x + 5)-(x + 1)] / (x (x + 2))
Expand the bracket
=[x² +5x - x -1] / [x (x + 2)]
=(x² - 4x - 1) / [x (x + 2)]
= (x² - 4x - 1)/ [x (x + 2)]
= (x² - 4x - 1) / [x² + 2x]
