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alekssr [168]
2 years ago
10

Solve the system by substitution. 4x + 2y = 10 x − y = 13

Mathematics
2 answers:
Rainbow [258]2 years ago
8 0

answer are attached in pic hope it's helpful for you

value of x =6

y =7

ziro4ka [17]2 years ago
5 0

Answer:

\boxed{\tt (6,-7)}

Step-by-step explanation:

\tt 4x + 2y = 10\\x - y = 13

First Let's solve for x in x-y=13

\tt x-y=13

Add y to both sides:

\tt x=13+y

Now, Substitute x=13+y into 4x+2y=10:-

Let x=13+y

\tt 4(13+y)+2y=10

\tt 52+6y=10

Now, let's solve for y in 52+6y=10:-

\tt 52+6y=10

Subtract 52 from both sides:-

\tt 6y=10-52

\tt 6y=-42

Divide both sides by 6:-

\tt \cfrac{6y}{6} =\cfrac{-42}{6}

\boxed{\tt y=-7}

Now, we'll Substitute y=-7 into to x=13+y to find "x":-

\tt x=13+y

\tt x=13+-7

\boxed{\tt x=6}

___

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Answer:

\frac{y}{x^2}=\sin x+\pi

Step-by-step explanation:

Consider linear differential equation \frac{\mathrm{d} y}{\mathrm{d} x}+yp(x)=q(x)

It's solution is of form y\,I.F=\int I.F\,q(x)\,dx where I.F is integrating factor given by I.F=e^{\int p(x)\,dx}.

Given: \frac{1}{x}\frac{\mathrm{d} y}{\mathrm{d} x}-\frac{2y}{x^2}=x\cos x

We can write this equation as \frac{\mathrm{d} y}{\mathrm{d} x}-\frac{2y}{x}=x^2\cos x

On comparing this equation with \frac{\mathrm{d} y}{\mathrm{d} x}+yp(x)=q(x), we get p(x)=\frac{-2}{x}\,\,,\,\,q(x)=x^2\cos x

I.F = e^{\int p(x)\,dx}=e^{\int \frac{-2}{x}\,dx}=e^{-2\ln x}=e^{\ln x^{-2}}=\frac{1}{x^2}      { formula used: \ln a^b=b\ln a }

we get solution as follows:

\frac{y}{x^2}=\int \frac{1}{x^2}x^2\cos x\,dx\\\frac{y}{x^2}=\int \cos x\,dx\\\\\frac{y}{x^2}=\sin x+C

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Applying condition:y(\pi)=\pi^2

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4 years ago
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