Answer:
The most Kyle can invest at 4% is $10000.
Step-by-step explanation:
To Determine:
What is the most he can invest at 4% and still be guaranteed at least $700 in interest per year?
Information Fetching and Solution Steps:
As
So,
If Kyle has to invest $16,000. then the following equations can be established from the given data
[tex]x + y = 16,000[/tex]
[tex].04x + .05y = 700[/tex]
Multiplying the second equation by 100, we get,
[tex]4x+5y = 70000[/tex]
Multiplying the 1st equation i.e. [tex]x + y = 16,000[/tex] by 5
[tex]5x + 5y = 80000[/tex]
So, the equations become
[tex]\begin{bmatrix}5x+5y=80000\\ 4x+5y=70000\end{bmatrix}[/tex]
[tex]\mathrm{Isolate}\:x\:\mathrm{for}\:5x+5y=80000:\quad x=16000-y[/tex]
As
[tex]5x+5y=80000[/tex]
[tex]5x+5y-5y=80000-5y[/tex]
[tex]5x=80000-5y[/tex]
[tex]\frac{5x}{5}=\frac{80000}{5}-\frac{5y}{5}[/tex]
[tex]x=16000-y[/tex]
[tex]\mathrm{Subsititute\:}x=16000-y[/tex]
[tex]\begin{bmatrix}4\left(16000-y\right)+5y=70000\end{bmatrix}[/tex]
[tex]\mathrm{Isolate}\:y\:\mathrm{for}\:4\left(16000-y\right)+5y=70000:\quad y=6000[/tex]
As
[tex]4\left(16000-y\right)+5y=70000[/tex]
[tex]64000-4y+5y=70000[/tex]
[tex]64000+y=70000[/tex]
[tex]y=6000[/tex]
[tex]\mathrm{For\:}x=16000-y[/tex]
[tex]\mathrm{Subsititute\:}y=6000[/tex]
[tex]x=16000-6000[/tex]
[tex]x=10000[/tex] $
Therefore, the most Kyle can invest at 4% is $10000.
Keywords: interest rate, equation
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