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Liono4ka [1.6K]

2 years ago

13

¿A cuál ecuación corresponde la expresión x=3?

2 answers:

vova2212 [387]2 years ago

7 0

Answer:

Estas hablando espanol? Perdon, mio es muy malo. Piensa que la responder correcta es -x=-3.

Step-by-step explanation:

Bas_tet [7]2 years ago

6 0

Answer:

(8x − 2x ) + (−x − 2) = (5x − 3) − (4 −x ) a) Ecuación de Primer Grado. b)

Step-by-step explanation:

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Is there anyone that can help me with this problem?

saul85 [17]

Answer:

your answer will be option 2

None of the choices are correct.

Step-by-step explanation:

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6 0

2 years ago

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kirza4 [7]

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6 0

3 years ago

Solve for x:<br> – 2x – 5 = x - 17<br> PLEASE HELP!

mariarad [96]

Answer:

x=4

Step-by-step explanation:

-2x-5-x=-17

-3x-5-x=17

-3x=-17+5

-3x=-12

x=-12÷-3

x=4

make me as brilliant answer

5 0

2 years ago

Solve the following equations.<br> log2(x^2 − 16) − log^2(x − 4) = 1

Alenkasestr [34]

Answer:

$x=\frac{4*(2+e)}{e-2}$

Step-by-step explanation:

Let's rewrite the left side keeping in mind the next propierties:

$log(\frac{1}{x} )=-log(x)$

$log(x*y)=log(x)+log(y)$

Therefore:

$log(2*(x^{2} -16))+log(\frac{1}{(x-4)^{2} })=1\\ log(\frac{2*(x^{2} -16)}{(x-4)^{2}})=1$

Now, cancel logarithms by taking exp of both sides:

$e^{log(\frac{2*(x^{2} -16)}{(x-4)^{2}})} =e^{1} \\\frac{2*(x^{2} -16)}{(x-4)^{2}}=e$

Multiply both sides by $(x-4)^{2}$ and using distributive propierty:

$2x^{2} -32=16e-8ex+ex^{2}$

Substract $16e-8ex+ex^{2}$ from both sides and factoring:

$-(x-4)*(-8-4e-2x+ex)=0$

Multiply both sides by -1:

$(x-4)*(-8-4e-2x+ex)=0$

Split into two equations:

$x-4=0\hspace{3}or\hspace{3}-8-4e-2x+ex=0$

Solving for $x-4=0$

Add 4 to both sides:

$x=4$

Solving for $-8-4e-2x+ex=0$

Collect in terms of x and add $4e+8$ to both sides:

$x(e-2)=4e+8$

Divide both sides by e-2:

$x=\frac{4*(2+e)}{e-2}$

The solutions are:

$x=4\hspace{3}or\hspace{3}x=\frac{4*(2+e)}{e-2}$

If we evaluate x=4 in the original equation:

$log(0)-log(0)=1$

This is an absurd because log (x) is undefined for $x\leq 0$

If we evaluate $x=\frac{4*(2+e)}{e-2}$ in the original equation:

$log(2*((\frac{4e+8}{e-2})^2-16))-log((\frac{4e+8}{e-2}-4)^2)=1$

Which is correct, therefore the solution is:

$x=\frac{4*(2+e)}{e-2}$

6 0

3 years ago

Solve each system by substitution 2x + y = 20 and 6x - 5y = 12

ivolga24 [154]

$\huge\mathbf{Answer࿐}$

$\underline{ \underline{ \bf{step} - 1}}$

Taking first equation,

$2x + y = 20$

$y = 20 - 2x$

$\underline{ \underline{ \bf{step} - 2}}$

plugging the value of y in second equation,

$6x - 5y = 12$

$6x - 5(20 - 2x) = 12$

$6x - 100 + 10x = 12$

$16x = 12 + 100$

$x = \dfrac{112}{16}$

$x = 7$

$\underline{ \underline{ \bf{step} -3}}$

plugging the value of x in first equation,

$2x + y = 20$

$(2 \times 7) + y = 20$

$14 + y = 20$

$y = 20 - 14$

$y = 6$

<h2>Conclusion :</h2>

$x = 7$
$y = 6$

____________________________

$\mathrm{ \#TeeNForeveR}$

6 0

2 years ago