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geniusboy [140]
2 years ago
10

On your own paper, solve the system of equations using elimination and identify the solution. Always list your answer alphabetic

ally in order pairs.
5e + 4f=9
4e+5f=9

A.(2,2)
B.(1,1)
C.(,1,-1(
D.(-2,-2)
E.IMS
F,NS
Mathematics
1 answer:
Oksana_A [137]2 years ago
4 0
5e + 4f = 9....multiply by -4
4e + 5f = 9 ...multiply by 5
------------------
-20e - 16f = -36 (result of multiplying by -4)
20e + 25f = 45 (result of multiplying by 5)
----------------add
9f = 9
f = 9/9
f = 1

5e + 4f = 9
5e + 4(1) = 9
5e + 4 = 9
5e = 9 - 4
5e = 5
e = 5/5
e = 1

solution is (1,1)


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The potential solutions of log_4x+log_4(x+6)=2 are 2 and -8.

<h3>Properties of Logarithms</h3>

From the properties of logarithms, you can rewrite logarithmic expressions.

The main properties are:

  • Product Rule for Logarithms - log_{b}(a*c)=log_{b}a+log_{b}c
  • Quotient Rule for Logarithms - log_{b}(\frac{a}{c} )=log_{b}a-log_{b}c
  • Power Rule for Logarithms - log_{b}(a^c)=c*log_{b}a

The exercise asks the potential solutions for  log_4x+log_4(x+6)=2. In this expression you can apply the Product Rule for Logarithms.

                                  log_4x+log_4(x+6)=2\\ \\ x*(x+6)=4^2\\ \\ x^2+6x=16\\ \\ x^2+6x-16=0

Now you should solve the quadratic equation.

 

 Δ=b^2-4ac=36-4*1*(-16)=36+64=100. Thus, x will be x_{1,\:2}=\frac{-6\pm \:\sqrt{100} }{2\cdot \:1}=\frac{-6\pm \:10}{2}. Then:

x_1=\frac{-6+10}{2}=\frac{4}{2} =2\\ \\ \:x_2=\frac{-6-10}{2}=\frac{-16}{2} =-8

The potential solutions  are 2 and -8.

Read more about the properties of logarithms here:

brainly.com/question/14868849

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