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exis [7]
3 years ago
12

Solve the system of equations 7x+10y=36 y=2x+9

Mathematics
1 answer:
Damm [24]3 years ago
5 0

Answer:

x = -2

y = 5

Step-by-step explanation:

Given two equations

1. 7x + 10y = 36

2. y = 2x + 9

Substitute 2x + 9 for y in equation 1.

We have

7x + 10y = 36

7x + 10(2x + 9) = 36

Distribute 10 into (2x + 9)

We have

7x + 10 x 2x + 10 x 9 = 36

7x + 20x + 90 = 36

27x + 90 = 36

Subtract 90 from both sides to eliminate 90 on the left side

27x + 90 - 90 = 36 - 90

27x = -54

Divide both sides by 27 to isolate x

27x/27 = -54/27

x = -2

Now substitute -2 into either equation to get y

Using equation 2, we have

y = 2x + 9

= 2 x -2 + 9

= -4 + 9

= 5

x = -2

y = 5

Check

7(-2) + 10(5) = 36

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Step-by-step explanation:

Given the geometric sequence

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A geometric sequence has a constant ratio and is defined by

a_n=a_1\cdot r^{n-1}

\mathrm{Compute\:the\:ratios\:of\:all\:the\:adjacent\:terms}:\quad \:r=\frac{a_{n+1}}{a_n}

\frac{6}{8}=\frac{3}{4},\:\quad \frac{4.5}{6}=\frac{3}{4}

\mathrm{The\:ratio\:of\:all\:the\:adjacent\:terms\:is\:the\:same\:and\:equal\:to}

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\mathrm{The\:first\:element\:of\:the\:sequence\:is}

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\mathrm{Therefore,\:the\:}n\mathrm{th\:term\:is\:computed\:by}\:

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a_1\frac{1-r^n}{1-r}

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=8\cdot \frac{1-\left(\frac{3}{4}\right)^{25}}{1-\frac{3}{4}}

\mathrm{Multiply\:fractions}:\quad \:a\cdot \frac{b}{c}=\frac{a\:\cdot \:b}{c}

=\frac{\left(1-\left(\frac{3}{4}\right)^{25}\right)\cdot \:8}{1-\frac{3}{4}}

=\frac{8\left(-\left(\frac{3}{4}\right)^{25}+1\right)}{\frac{1}{4}}

\mathrm{Apply\:exponent\:rule}:\quad \left(\frac{a}{b}\right)^c=\frac{a^c}{b^c}

=\frac{8\left(-\frac{3^{25}}{4^{25}}+1\right)}{\frac{1}{4}}

\mathrm{Apply\:the\:fraction\:rule}:\quad \frac{a}{\frac{b}{c}}=\frac{a\cdot \:c}{b}

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=\frac{32\cdot \frac{4^{25}-3^{25}}{4^{25}}}{1}               ∵ \mathrm{Join}\:1-\frac{3^{25}}{4^{25}}:\quad \frac{4^{25}-3^{25}}{4^{25}}

=32\cdot \frac{4^{25}-3^{25}}{4^{25}}

=\frac{\left(4^{25}-3^{25}\right)\cdot \:32}{4^{25}}

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\mathrm{Apply\:the\:fraction\:rule}:\quad \frac{a\pm \:b}{c}=\frac{a}{c}\pm \frac{b}{c}

=\frac{4^{25}}{2^{45}}-\frac{3^{25}}{2^{45}}      

=32-\frac{3^{25}}{2^{45}}            ∵  \frac{4^{25}}{2^{45}}=32

=32-0.024        ∵  \frac{3^{25}}{2^{45}}=0.024

=31.98            

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