Answer:
(1, -2)
General Formulas and Concepts:
<u>Pre-Algebra</u>
Order of Operations: BPEMDAS
- Brackets
- Parenthesis
- Exponents
- Multiplication
- Division
- Addition
- Subtraction
Equality Properties
<u>Algebra I</u>
- Solving systems of equations using substitution/elimination
Step-by-step explanation:
<u>Step 1: Define Systems</u>
4x + y = 2
y = x - 3
<u>Step 2: Solve for </u><em><u>x</u></em>
<em>Substitution</em>
- Substitute in <em>y</em>: 4x + (x - 3) = 2
- Combine like terms: 5x - 3 = 2
- Isolate <em>x</em> term: 5x = 5
- Isolate <em>x</em>: x = 1
<u>Step 3: Solve for </u><em><u>y</u></em>
- Define equation: y = x - 3
- Substitute in <em>x</em>: y = 1 - 3
- Subtract: y = -2
the function f(x) = 4(x+3) − 5
Lets find the inverse function f^-1(x)
step 1: Replace f(x) with y
y = 4(x+3) − 5
step 2: Replace x with y and y with x
x = 4(y+3) - 5
step 3: Solve for y
x = 4(y+3) - 5
x = 4y +12 -5
x= 4y + 7 (subtract 7 on both sides)
x - 7 = 4y (divide by 4 on both sides)
Given : x = 3
We plug in 3 for x in the inverse function
= -1
the inverse function when x = 3 is -1
Basically just add them all together like normal addition except there is a z attached to each number:
43z + 15z + 7z + 5z + 46z + 14z
58z+ 7z + 5z + 46z + 14z
65z + 5z + 46z + 14z
70z + 46z + 14z
116z + 14z
130z
Hope this helped!
