Algebra Problems and Solutions
In algebra problems, You must know how to:
1. Solve linear equations.
2. Simplify algebraic expressions.
3. Simplify the expression of absolute values.
4. Find the distance between two points.
5. Search x intercept from the equation graph.
6. Evaluating functions.
7. Find the slope of the line that passes through two points.
8. Find the slope of the line from the equation.
8. Find the equation of the line.
10. Solve equations with absolute values.
Algebraic Problems:
Problem 1: Solve the equation \begin{align} 5 (- 3x - 2) - (x - 3) = -4 (4x + 5) + 13 \end{align}
Solution 1: \begin{align} -15x-10 - x+3 &=-16x-20+13 \\ -15x-x-10+3 &= -16x-7 \\ -16x-7 &= -16x-7 \\ -16x+16x &= -7+7 \\ 0 &= 0 \end{align} The above statement is true for all values of x and therefore all real numbers are solutions to the given equation.
Problem 2: Simplify the expression $$2 (a -3) + 4b - 2 (a -b -3) + 5$$
Solution 2: \begin{align} & 2a-6+4b-2a+2b+6+5 \\ = & 2a-2a+4b +2b-6+6+5 \\ = & 6b+5 \end{align}
Problem 3: If x <2, simplify | x - 2 | - 4 | -6 |
Solution 3:
\begin{align} |x-2|-4|-6| &= |x-2|-4(6) \\ &= |x-2| - 24 \end{align}
if $x < 2$, then $|x-2|=-(x-2)=-x+2$. So, \begin{align}|x-2|-24=-x+2-24=-x-22 \end{align}
Problem 4: Find the distance between points (-4, -5) and (-1, -1).
Solution 4: \begin{align} d&= \sqrt{(-1-(-5))^2+(-1-(-4))^2} \\ &= \sqrt{(4)^2+(3)^2} \\ &= \sqrt{16+9} \\ &= \sqrt{25} \\ &= 5 \end{align}
Problem 5: Find x intercept from the equation graph. $$2x - 4y = 9$$
Solution 5:
Problem 6: Evaluation of $f (2) - f (1)$ if $f (x) = 6x +1$
Solution 6:
Problem 7: Find the slope of the line past point (-1, -1) and (2, 2).
Solution 7:
Problem 8: Find the slope of the line $5x - 5y = 7$
Solution 8:
Problem 9: Find the equation of the line passing point (-1, -1) and (-1, 2).
Solution 9:
Problem 10: Complete the equation $| - 2 x + 2 | - 3 = -3$
Solution 10:
Source of Problem: www.analyzemath.com/Algebra1/Algebra1.html
1. Solve linear equations.
2. Simplify algebraic expressions.
3. Simplify the expression of absolute values.
4. Find the distance between two points.
5. Search x intercept from the equation graph.
6. Evaluating functions.
7. Find the slope of the line that passes through two points.
8. Find the slope of the line from the equation.
8. Find the equation of the line.
10. Solve equations with absolute values.
Algebraic Problems:
Problem 1: Solve the equation \begin{align} 5 (- 3x - 2) - (x - 3) = -4 (4x + 5) + 13 \end{align}
Solution 1: \begin{align} -15x-10 - x+3 &=-16x-20+13 \\ -15x-x-10+3 &= -16x-7 \\ -16x-7 &= -16x-7 \\ -16x+16x &= -7+7 \\ 0 &= 0 \end{align} The above statement is true for all values of x and therefore all real numbers are solutions to the given equation.
Problem 2: Simplify the expression $$2 (a -3) + 4b - 2 (a -b -3) + 5$$
Solution 2: \begin{align} & 2a-6+4b-2a+2b+6+5 \\ = & 2a-2a+4b +2b-6+6+5 \\ = & 6b+5 \end{align}
Problem 3: If x <2, simplify | x - 2 | - 4 | -6 |
Solution 3:
\begin{align} |x-2|-4|-6| &= |x-2|-4(6) \\ &= |x-2| - 24 \end{align}
if $x < 2$, then $|x-2|=-(x-2)=-x+2$. So, \begin{align}|x-2|-24=-x+2-24=-x-22 \end{align}
Problem 4: Find the distance between points (-4, -5) and (-1, -1).
Solution 4: \begin{align} d&= \sqrt{(-1-(-5))^2+(-1-(-4))^2} \\ &= \sqrt{(4)^2+(3)^2} \\ &= \sqrt{16+9} \\ &= \sqrt{25} \\ &= 5 \end{align}
Problem 5: Find x intercept from the equation graph. $$2x - 4y = 9$$
Solution 5:
Problem 6: Evaluation of $f (2) - f (1)$ if $f (x) = 6x +1$
Solution 6:
Problem 7: Find the slope of the line past point (-1, -1) and (2, 2).
Solution 7:
Problem 8: Find the slope of the line $5x - 5y = 7$
Solution 8:
Problem 9: Find the equation of the line passing point (-1, -1) and (-1, 2).
Solution 9:
Problem 10: Complete the equation $| - 2 x + 2 | - 3 = -3$
Solution 10:
Source of Problem: www.analyzemath.com/Algebra1/Algebra1.html
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