23% is the answer i believe
With 433 dollars, you can buy 24 boxes of fixtures. When you buy 24 boxes, you will have 1 dollar left.
Answer:
No solutions.
General Formulas and Concepts:
<u>Pre-Algebra</u>
- Order of Operations: BPEMDAS
- Equality Properties
<u>Algebra I</u>
- Solving systems of equations using substitution/elimination
- Solving systems of equations by graphing
- Expanding
- Finding roots of a quadratic
- Standard Form: ax² + bx + c = 0
- Quadratic Formula:
![x=\frac{-b\pm\sqrt{b^2-4ac} }{2a}](https://tex.z-dn.net/?f=x%3D%5Cfrac%7B-b%5Cpm%5Csqrt%7Bb%5E2-4ac%7D%20%7D%7B2a%7D)
Step-by-step explanation:
<u>Step 1: Define systems</u>
2x - y = 9
4x² + 3y² - 2x + y = 16
<u>Step 2: Rewrite systems</u>
2x - y = 9
- Subtract 2x on both sides: -y = 9 - 2x
- Divide -1 on both sides: y = 2x - 9
<u>Step 3: Redefine systems</u>
y = 2x - 9
4x² + 3y² - 2x + y = 16
<u>Step 4: Solve for </u><em><u>x</u></em>
<em>Substitution</em>
- Substitute in <em>y</em>: 4x² + 3(2x - 9)² - 2x + (2x - 9) = 16
- Expand: 4x² + 3(4x² - 36x + 81) - 2x + (2x - 9) = 16
- Distribute 3: 4x² + 12x² - 108x + 243 - 2x + 2x - 9 = 16
- Combine like terms: 16x² - 108x + 234 = 16
- Factor GCF: 2(8x² - 54x + 117) = 16
- Divide 2 on both sides: 8x² - 54x + 117 = 8
- Subtract 8 on both sides: 8x² - 54x + 109 = 0
- Define variables: a = 8, b = -54, c = 109
- Resubstitute:
![x=\frac{54\pm\sqrt{(-54)^2-4(8)(109)} }{2(8)}](https://tex.z-dn.net/?f=x%3D%5Cfrac%7B54%5Cpm%5Csqrt%7B%28-54%29%5E2-4%288%29%28109%29%7D%20%7D%7B2%288%29%7D)
- Exponents:
![x=\frac{54\pm\sqrt{2916-4(8)(109)} }{2(8)}](https://tex.z-dn.net/?f=x%3D%5Cfrac%7B54%5Cpm%5Csqrt%7B2916-4%288%29%28109%29%7D%20%7D%7B2%288%29%7D)
- Multiply:
![x=\frac{54\pm\sqrt{2916-3488} }{16}](https://tex.z-dn.net/?f=x%3D%5Cfrac%7B54%5Cpm%5Csqrt%7B2916-3488%7D%20%7D%7B16%7D)
- Subtract:
![x=\frac{54\pm\sqrt{-572} }{16}](https://tex.z-dn.net/?f=x%3D%5Cfrac%7B54%5Cpm%5Csqrt%7B-572%7D%20%7D%7B16%7D)
Here we see that we start to delve into imaginary roots. Since on a real number plane, we do not have imaginary roots, there would be no solution to the systems of equations.
<u>Step 5: Graph systems</u>
<em>We can verify our results.</em>
Answer:
NOTES
Step-by-step explanation: