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3 years ago

PLEASE HELP ME ASAP!

Mathematics

2 answers:

AleksandrR [38]3 years ago

6 0

Ok so first we find the equation that equals one variable.

2y = -x + 9

3x - 6y = -15

We solve for y.

2y = -x + 9

y = -x/2 + 9/2

Then we plug in this y value into the other equation to keep only one variable so we can solve for it.

3x - 6y = -15

3(-x + 9/2) - 6y = -15

-3x + 27/2 - 6y = -15

-9y + 27/2 = -15

-9y = 3/2

-y = 3/18

y = -3/18

Then we plug in this numerical y-value into the first equation which we found out by solving an equation for y.

y = -x/2 + 9/2

-3/18 = -x/2 + 9/2

-84/18 = -x/2

-x = 9 1/3

x = -28/3

Your answer would be (-28/3, -3/18)

Hope this helps!

aalyn [17]3 years ago

6 0

(For this, I'll be using the substitution method.)

So firstly, subtract 9 on each side of the first equation. Your equation will look like this: $2y-9=-x$

Next, flip the signs around in the first equation. Your equation will look like this: $-2y+9=x$

Next, substitute x in the second equation for (-2y+9). From there you can solve for y. $3(-2y+9)-6y=15$

Firstly, foil 3(-2y+9). $-6y+27-6y=15$

Next, combine like terms: $-12y+27=15$

Subtract 27 on each side: $-12y=-12$

Lastly, divide -12 on each side, and your answer should be $y=1$

To solve for x, just substitute y for 1 in either equation. For this Ill be doing the first equation.

$2(1)=-x+9$

Multiply 2 and 1 $2=-x+9$

Subtract 9 on each side $-7=-x$

Lastly, flip the signs, and your answer should be $7=x$

In short, the answer is (7,1).

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$\displaystyle \begin{cases} \displaystyle {x} _{1} = - p \\ \displaystyle x _{2} = - q \end{cases}$

Step-by-step explanation:

we would like to solve the following equation for x:

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to do so isolate $\frac{1}{x}$ to right hand side and change its sign which yields:

$\displaystyle \frac{1}{p} + \frac{1}{q} = \frac{1}{p + q + x} - \frac{1}{x}$

simplify Substraction:

$\displaystyle \frac{1}{p} + \frac{1}{q} = \frac{x - (q + p + x)}{x(p + q + x)}$

get rid of only x:

$\displaystyle \frac{1}{p} + \frac{1}{q} = \frac{ - (q + p )}{x(p + q + x)}$

simplify addition of the left hand side:

$\displaystyle \frac{q + p}{pq} = \frac{ - (q + p )}{x(p + q + x)}$

divide both sides by q+p Which yields:

$\displaystyle \frac{1}{pq} = \frac{ -1}{x(p + q + x)}$

cross multiplication:

$\displaystyle x(p + q + x) = - pq$

distribute:

$\displaystyle xp + xq + {x}^{2} = - pq$

isolate -pq to the left hand side and change its sign:

$\displaystyle xp + xq + {x}^{2} + pq = 0$

rearrange it to standard form:

$\displaystyle {x}^{2} + xp + xq + pq = 0$

now notice we end up with a quadratic equation therefore to solve so we can consider factoring method to use so

factor out x:

$\displaystyle x( {x}^{} + p ) + xq + pq = 0$

factor out q:

$\displaystyle x( {x}^{} + p ) +q (x + p)= 0$

group:

$\displaystyle ( {x}^{} + p ) (x + q)= 0$

by Zero product property we obtain:

$\displaystyle \begin{cases} \displaystyle {x}^{} + p = 0 \\ \displaystyle x + q= 0 \end{cases}$

cancel out p from the first equation and q from the second equation which yields:

$\displaystyle \begin{cases} \displaystyle {x}^{} = - p \\ \displaystyle x = - q \end{cases}$

and we are done!

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