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marshall27 [118]
3 years ago
9

2x+8y=3 and 4x-3y=2 value of x and y.. plzz help me​

Mathematics
1 answer:
lesya692 [45]3 years ago
6 0

Answer:  

<u>Set a system of equations:</u>

\left \{ {{2x+8y=3} \atop {4x-3y=2}} \right.

<u>Multiply the whole upper equation by 2 and the whole bottom equation by 1 so their x-value equals:</u>

\left \{ {{2(2)x+8(2)y=3(2)} \atop {4(1)x-3(1)y=2(1)}} \right.\\\\=\left \{ {{4x+16y=6} \atop {4x-3y=2}} \right.

<u>Subtract the upper equation by the bottom equation & solve for y:</u>

(4x-4x)+(16y-(-3y))=6-2\\\\19y=4\\\\y=\frac{4}{19}

<u>Substitute in the y-value to a equation to find x:</u>

2x+8y=3\\\\2x+8(\frac{4}{19} )=3\\\\2x=3-\frac{32}{19} =\frac{57}{19} -\frac{32}{19} =\frac{25}{19} \\\\x=\frac{\frac{25}{19}}{2} =\frac{25}{19}*\frac{1}{2} =\frac{25}{38}

<u>Therefore, the answer would be:</u>

<u />\left \{ {{y=\frac{4}{19} } \atop {x=\frac{25}{38} }} \right.

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Given:

The equation is,

2\log _3x-\log _3(x-2)=2

Explanation:

Simplify the equation by using logarthimic property.

\begin{gathered} 2\log _3x-\log _3(x-2)=2 \\ \log _3x^2-\log _3(x-2)=2_{}\text{      \lbrack{}log(a)-log(b) = log(a/b)\rbrack} \\ \log _3\lbrack\frac{x^2}{x-2}\rbrack=2 \end{gathered}

Simplify further.

\begin{gathered} \log _3\lbrack\frac{x^2}{x-2}\rbrack=2 \\ \frac{x^2}{x-2}=3^2 \\ x^2=9(x-2) \\ x^2-9x+18=0 \end{gathered}

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\begin{gathered} x^2-6x-3x+18=0 \\ x(x-6)-3(x-6)=0 \\ (x-6)(x-3)=0 \end{gathered}

From the above equation (x - 6) = 0 or (x - 3) = 0.

For (x - 6) = 0,

\begin{gathered} x-6=0 \\ x=6 \end{gathered}

For (x - 3) = 0,

\begin{gathered} x-3=0 \\ x=3 \end{gathered}

The values of x from solving the equations are x = 3 and x = 6.

Substitute the values of x in the equation to check answers are valid or not.

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\begin{gathered} 2\log _3(3^{})-\log _3(3-2)=2 \\ 2\log _33-\log _31=2 \\ 2\cdot1-0=2 \\ 2=2 \end{gathered}

Equation satisfy for x = 3. So x = 3 is valid value of x.

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\begin{gathered} 2\log _36-\log _3(6-2)=2 \\ 2\log _36-\log _34=2 \\ \log _3(6^2)-\log _34=2 \\ \log _3(\frac{36}{4})=2 \\ \log _39=2 \\ \log _3(3^2)=2 \\ 2\log _33=2 \\ 2=2 \end{gathered}

Equation satifies for x = 6.

Thus values of x for equation are x = 3 and x = 6.

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