Im pretty sure It would be a decrease
Answer:
x = 13
General Formulas and Concepts:
<u>Pre-Algebra</u>
Order of Operations: BPEMDAS
- Brackets
- Parenthesis
- Exponents
- Multiplication
- Division
- Addition
- Subtraction
Equality Properties
- Multiplication Property of Equality
- Division Property of Equality
- Addition Property of Equality
- Subtract Property of Equality<u>
</u>
<u>Algebra I</u>
<u>Trigonometry</u>
[Right Triangles Only] Pythagorean Theorem: a² + b² = c²
- a is a leg
- b is another leg
- c is the hypotenuse<u>
</u>
Step-by-step explanation:
<u>Step 1: Identify</u>
<em>a</em> = 12
<em>b</em> = 5
<em>c</em> = x
<u>Step 2: Solve for </u><em><u>x</u></em>
- Substitute in variables [PT]: 12² + 5² = x²
- Evaluate exponents: 144 + 25 = x²
- Add: 169 = x²
- [Equality] Square root both sides: ±13 = x
- Rewrite: x = ±13
Since we are dealing with positive numbers, we can disregard the negative root.
∴ x = 13
<span>A linear equation in one variable has a single unknown quantity called a variable represented by a letter. Eg: ‘x’, where ‘x’ is always to the power of 1. This means there is no ‘ x² ’ or ‘ x³ ’ in the equation.The process of finding out the variable value that makes the equation true is called ‘solving’ the equation.An equation is a statement that two quantities are equivalent.For example, this linear equation: x<span> + 1 = 4 </span>means that when we add 1 to the unknown value, ‘x’, the answer is equal to 4.To solve linear equations, you add, subtract, multiply and divide both sides of the equation by numbers and variables, so that you end up with a single variable on one side and a single number on the other side. As long as you always do the same thing to BOTH sides of the equation, and do the operations in the correct order, you will get to the solution.</span><span><span>For this example, we only need to subtract 1 from both sides of the equation in order to isolate 'x' and solve the equation:x<span> + 1 </span>-<span> 1 = 4 </span>-<span> 1</span>Now simplifying both sides we have:x<span> + 0 = 3</span>So:</span><span>x<span> = 3</span></span></span><span>With some practice you will easily recognise what operations are required to solve an equation.Here are possible ways of solving a variety of linear equation types.<span>Example 1, Solve for ‘x’ :</span>x<span> + 1 = </span>-31. Subtract 1 from both sides:x<span> + 1 </span>-<span> 1 = </span>-<span>3 </span>-<span> 1</span>2. Simplify both sides:x<span> = </span>-4<span>Example 2, Solve for ‘x’ :</span>-<span>2x = 12</span>1. Divide both sides by -2:2. Simplify both sides:x<span> = </span>-6<span>Example 3, Solve for ‘x’ :</span>1. Multiply both sides by 3:2. Simplify both sides:<span>x = </span>-6<span>Example 4, Solve for ‘x’ :</span><span>2x + 1 = </span>-171. Subtract 1 from both sides:<span>2x + 1 </span>-<span> 1 = </span>-<span>17 </span>-<span> 1</span>2. Simplify both sides:<span>2x = </span>-183. Divide both sides by 2:4. Simplify both sides:<span>x = </span>-9<span>Example 5, Solve for ‘x’ :</span>1. Multiply both sides by 9:2. Simplify both sides:<span>3x = 36</span>3. Divide both sides by 3:4. Simplify both sides:x = 12<span>Example 6, Solve for ‘x’ :</span> 1. Multiply both sides by 3: 2. Simplify both sides:<span> x + 1 = 21</span> 3. Subtract 1 from both sides:<span> x + 1 </span>-<span> 1 = 21 </span>-<span> 1</span> 4. Simplify both sides:x = 20<span>Example 7, Solve for ‘x’ :</span><span>7(x </span>-<span> 1) = 21</span>1. Divide both sides by 7:2. Simplify both sides:<span>x </span>-<span> 1 = 3</span>3. Add 1 to both sides:<span>x </span>-<span> 1 + 1 = 3 + 1</span>4. Simplify both sides:x = 4<span>Example 8, Solve for ‘x’ :</span>1. Multiply both sides by 5:2. Simplify both sides:<span>3(x </span>-<span> 1) = 30</span>3. Divide both sides by 3:4. Simplify both sides:<span>x </span>-<span> 1 = 10</span>5. Add 1 to both sides:<span>x </span>-<span> 1 + 1 = 10 + 1</span>6. Simplify both sides:x<span> = 11</span><span>Example 9, Solve for ‘x’ :</span><span>5x + 2 = 2x + 17</span>1. Subtract 2x from both sides:<span>5x + 2 </span>-<span> 2x = 2x + 17 </span>-<span> 2x</span>2. Simplify both sides:<span>3x + 2 = 17</span>3. Subtract 2 from both sides:<span>3x + 2 </span>-<span> 2 = 17 </span>-<span> 2</span>4. Simplify both sides:<span>3x = 15</span>5. Divide both sides by 3:6. Simplify both sides:x = 5<span>Example 10, Solve for ‘x’ :</span><span>5(x </span>-<span> 4) = 3x + 2</span>1. Expand brackets:<span>5x </span>-<span> 20 = 3x + 2</span>2. Subtract 3x from both sides:<span>5x </span>-<span> 20 </span>-<span> 3x = 3x + 2 </span>-<span> 3x</span>3. Simplify both sides:<span>2x </span>-<span> 20 = 2</span>4. Add 20 to both sides:<span>2x </span>-<span> 20 + 20 = 2 + 20</span>5. Simplify both sides:<span>2x = 22</span>6. Divide both sides by 2:7. Simplify both sides:x <span>= 11</span></span>
10 6 12 5 30 2 15 4 60 1 and others if not whole numbers
