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3 years ago

Solve and verify please solve

Mathematics

1 answer:

RUDIKE [14]3 years ago

5 0

Answer:

Below in bold.

Step-by-step explanation:

I'll solve a few of them to illustrate the method:

a) 5x - (3x - 1) = x - 4

The aim is to get all the 'x' terms to one sides of the '=' and the numbers to the other side:

Distribute the negative over the 3x - 1 in the parentheses:

5x - 3x + 1 = x - 4 ( Note that -* -1 = + 1)

Now subtract 1 from both sides:

5x - 3x + 1 - 1 = x - 4 - 1

5x - 3x = x - 5

Now subtract x from both sides:

5x - 3x - x = x - x - 5.

2x - x = -5

x = -5.

(b) 3/5 x + 1 = 2/5 - 3x

We first get rid of the fractions by multiplying each term by 5:

5 * 3/5 x + 1*5 = 5 * 2/5 - 5*3x

3x + 5 = 2 - 15x

3x + 15x + 5 = 2 - 15x + 15x (adding 15x to both sides)

18x + 5 = 2 Subtract 5 from both sides:

18x + 5 - 5 = 2 - 5

18x = -3

Now we divide both sides by 18:

x = -3/18 = -1/6.

g) 0.3(6 + m) = 0.4(8 - m)

0.3 * 6 + 0.3* m = 0.4*8 + 0.4 * -m ( Distributing the 0.3 and 0.4 over the parentheses)

1.8 + 0.3m = 3.2 - 0.4m ( recall that + * - = -)

1.8 - 1.8 + 0.3m = 3.2 - 1.8 - 0.4m

0.3m = 1.4 - 0.4m

0.3m + 0.4m = 1.4 - 0.4m + 0.4m

0.7m = 1.4

m = 1/4/0.7

m = 2.

I hope this helps.

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This is an ellipse. The equation is:

$\frac{(x-1)^2}{3^2}+\frac{(y+4)^2}{4^2}=1$

EXPLANATION

We have to complete the square for each variable. To do so, we have to take the first two terms and compare them with the perfect binomial squared formula,

$(a+b)^2=a^2+2ab+b^2$

For x we have to take 16x² and -32x. Since the coefficient of x is not 1, first, we have to factor out the coefficient 16,

$16x^2-32x=16(x^2-2x)$

Now, the first term of the expanded binomial would be x and the second term -2x. Thus, the binomial is,

$(x-1)^2=x^2-2x+1$

To maintain the equation, we have to subtract 1,

$16(x^2-2x+1-1)=16((x-1)^2-1)=16(x-1)^2-16$

Now, we replace (16x² - 32x) from the given equation by this equivalent expression,

$16(x-1)^2-16+9y^2+72y+16=0$

The next step is to do the same for y. We have the terms 9y² + 72y. Again, since the coefficient of y² is not 1, we factor out the coefficient 9,

$9y^2+72y=9(y^2+8y)$

Following the same reasoning as before, we have that the perfect binomial squared is,

$(y+4)^2=y^2+8y+16$

Remember to subtract the independent term to maintain the equation,

$9(y^2+8y)=9(y^2+8y+16-16)=9((y+4)^2-16)=9(y+4)^2-144$

And now, as we did for x, replace the two terms (9y² + 72y) with this equivalent expression in the equation,

$16(x-1)^2-16+9(y+4)^2-144+16=0$

Add like terms,

$\begin{gathered} 16(x-1)^2+9(y+4)^2+(-16-144+16)=0 \\ 16(x-1)^2+9(y+4)^2-144=0 \end{gathered}$

Add 144 to both sides,

$\begin{gathered} 16(x-1)^2+9(y+4)^2-144+144=0+144 \\ 16(x-1)^2+9(y+4)^2=144 \end{gathered}$

As we can see, this is the equation of an ellipse. Its standard form is,

$\frac{(x-h)^2}{a^2}+\frac{(y-k)^2}{b^2}=1$

So the next step is to divide both sides by 144 and also write the coefficients as fractions in the denominator,

$\begin{gathered} \frac{16(x-1)^2}{144}+\frac{9(y+4)^2}{144}=\frac{144}{144} \\ \\ \frac{(x-1)^2}{\frac{144}{16}}+\frac{(y+4)^2}{\frac{144}{9}}=1 \end{gathered}$

Finally, we have to write the denominators as perfect squares, so we identify the values of a and b. 144 is 12², 16 is 4² and 9 is 3²,

$\frac{(x-1)^2}{(\frac{12}{4})^2}+\frac{(y+4)^2}{(\frac{12}{3})^2}=1$

Note that we can simplify a and b,

$\frac{12}{4}=3\text{ and }\frac{12}{3}=4$

Hence, the equation of the ellipse is,

$\frac{(x-1)^2}{3^2}+\frac{(y+4)^2}{4^2}=1$

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