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3 years ago

I have to solve the system of linear equations by elimination. i also have to check my solution.

Mathematics

1 answer:

Travka [436]3 years ago

6 0

Answer:

$x=1$ , $y=6$

Step-by-step explanation:

Give.

$x+2y=13$ ----------(1)

$-x+y=5$ ------------(2)

Solve this linear equation by elimination method.

Add equation 1 and 2

$x+2y=13$

$-x+y=5$

-----------------------------

$0+3y=18$

$3y=18$

$y=\frac{18}{3} \\y=6$

Put y value in equation 2

$-x+6=5$

$-x=5-6$

$-x=-1$

$x=1$

$x=1$ , $y=6$

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4 years ago

g find the 2 components of vector b = 2i + j - 3k, one parallel to a = 3i - j and another one perpendicular to a

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Step-by-step explanation:

Let be $\vec b = 2\,i+j-3\,k$ and $\vec a = 3\,i-j$ , the component of $\vec b$ parallel to $\vec a$ is calculated by the following expression:

$\vec b_{\parallel} = (\vec b \bullet \hat{a}) \cdot \hat{a}$

Where $\hat{a}$ is the unit vector of $\vec a$ , dimensionless and $\bullet$ is the operator of scalar product.

The unit vector of $\vec a$ is:

$\hat{a} = \frac{\vec {a}}{\|\vec a\|}$

Where $\|\vec {a}\|$ is the norm of $\vec a$ , whose value is determined by Pythagorean Theorem.

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$\|\vec {a}\| = \sqrt{3^{2}+(-1)^{2}+0^{2}}$

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$\hat{a} = \frac{1}{\sqrt{10}} \cdot (3\,i-j)$

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$\vec{b}\bullet \hat{a} = (2)\cdot \left(\frac{3}{\sqrt{10}} \right)+(1)\cdot \left(-\frac{1}{\sqrt{10}} \right)+(-3)\cdot \left(0\right)$

$\vec b \bullet \hat{a} = \frac{5}{\sqrt{10}}$

$\vec b_{\parallel} = \frac{5}{\sqrt{10}}\cdot \left(\frac{3}{\sqrt{10}}\,i-\frac{1}{\sqrt{10}}\,j \right)$

$\vec {b}_{\parallel} = \frac{3}{2}\,i-\frac{1}{2}\,j$

Now, the component of $\vec {b}$ perpendicular to $\vec{a}$ is found by vector subtraction:

$\vec{b}_{\perp} = \vec {b}-\vec {b}_{\parallel}$

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$\vec{b}_{\perp} = (2\,i+j-3\,k)-\left(\frac{3}{2}\,i-\frac{1}{2}\,j \right)$

$\vec b _{\perp} = \frac{1}{2}\,i+\frac{3}{2}\,j-3\,k$

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