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3 years ago

Solve by substitution: -5x + y = -2 -3x + 6y = -12 please explain

Mathematics

1 answer:

fomenos3 years ago

4 0

First we pick the equation. Let's say we pick first one. From it we express y.

$-5x+y=-2\Longrightarrow y=-2+5x$

Then we use this y and plug it in instead of y in the second equation.

$-3x+6(-2+5x)=-12$

Now just solve for x.

$-3x-12+30x=-12 \\-3x=-30x \\\underline{x=10}$

We plug this x in the first equation.

$-5\cdot10+y=-2$

And solve for y.

$-50+y=-2 \\\underline{y=48}$

So the solution to the equation is geometrically a point P which lies on the intersection of the two lines.

$\boxed{P(10,48)}$

Hope this helps.

r3t40

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Use a proof by contradiction to show that the square root of 3 is national You may use the following fact: For any integer kirke

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1. Let us proof that √3 is an irrational number, using reductio ad absurdum. Assume that $\sqrt{3}=\frac{m}{n}$ where $m$ and $n$ are non negative integers, and the fraction $\frac{m}{n}$ is irreducible, i.e., the numbers $m$ and $n$ have no common factors.

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Thus, 3 divides $n^2$ and necessarily 3 must divide $n$ . Hence, $n=3q$ where $q$ is a non negative integer.

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$\frac{m}{n} = \frac{3p}{3q} = \frac{p}{q}$ .

The above equality means that the fraction $\frac{m}{n}$ is reducible, what contradicts our initial assumption. So, $\sqrt{3}$ is irrational.

2. Let us prove now that the multiplication of an integer and a rational number is a rational number. So, $r\in\mathbb{Q}$ , which is equivalent to say that $r=\frac{m}{n}$ where $m$ and $n$ are non negative integers. Also, assume that $k\in\mathbb{Z}$ . So, we want to prove that $k\cdot r\in\mathbb{Z}$ . Recall that an integer $k$ can be written as

$k=\frac{k}{1}$ .

Then,

$k\cdot r = \frac{k}{1}\frac{m}{n} = \frac{mk}{n}$ .

Notice that the product $mk$ is an integer. Thus, the fraction $\frac{mk}{n}$ is a rational number. Therefore, $k\cdot r\in\mathbb{Q}$ .

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Write $q=x+p$ and let us suppose that $q$ is a rational number. So, we get that

$x=q-p$ .

But the subtraction or addition of two rational numbers is rational too. Then, the number $x$ must be rational too, which is a clear contradiction with our hypothesis. Therefore, $x+p$ is irrational.

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3 years ago