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Algebra

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RUNNING HEAD: ALGEBRAIC EQUATION
Algebraic Equation
Name
Institutional Affiliation

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ALGEBRAIC EQUATION 2
Q. If x²+4y²=40 and xy=6, what will be the value of x+2y? Please explain how you arrived at
your answer.
Solution:
i.
  
ii. 
Make x the subject of the formula of ii,
Replace x in the equation ii into i,
  


 

Divide all terms with ,

 



Factor 4 out of the equation,


 
Dealing with this equation

 ,
The 1
st
term is, y
4
its coefficient =1.
The 2
nd
term is, -10y
2
its coefficient = -10.

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RUNNING HEAD: ALGEBRAIC EQUATION Algebraic Equation Name Institutional Affiliation ALGEBRAIC EQUATION 2 Q. If x²+4y²=40 and xy=6, what will be the value of x+2y? Please explain how you arrived at your answer. Solution: i. 𝑥 2 + 4𝑦² = 40 ii. 𝑥𝑦 = 6 Make x the subject of the formula of ii, 6 𝑥=𝑦 Replace x in the equation ii into i, 6 (𝑦)2 + 4𝑦² = 40 36 𝑦² + 4𝑦 2 = 40 Divide all terms with 𝑦², 4𝑦 4 + 36 = 40𝑦² 4𝑦 4 − 40𝑦 2 + 36 = 0 Factor 4 out of the equation, 4(𝑦 4 − 10𝑦 2 + 9) = 0 Dealing with this equation 𝑦 4 − 10𝑦 2 + 9, The 1st term is, y4 its coefficient =1. The 2nd term is, -10y2 its coefficient = -10. ALGEBRAIC EQUATION The 3rd term, "the constant", = +9 iii. Multiply the coefficient of the first term by the constant 1 * 9 = 9 iv. Find two factors of 9 whose sum equals the coefficient of the middle term, which is 10. And it is -9 + -1 = -10, v. Rewrite the polynomial splitting the 2nd term using the two factors found in step 2 above, 9 and -1 and it becomes y4 − 9𝑦 2 − 1𝑦 2 − 9, vi. Add up 1st two like terms by pulling out common factor: vii. 𝑦 2 ∗ (𝑦 2 − 9) and 1 ∗ (𝑦 2 − 9) viii. Adding up all the four terms we get the following factorization, ix. (𝑦 2 − 1)(𝑦 2 − 9) x. Now we get the factor of different squares, From difference of two perfect squares, 𝑋 2 − 𝑌 2 = (𝑋 + 𝑌) ∗ (𝑋 − 𝑌) 𝑦 2 − 1 = (𝑦 + 1 ...
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