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3 years ago

Solve by substitutionx+3y=5 -2x-4=-5

Mathematics

1 answer:

likoan [24]3 years ago

3 0

Answer:

x = 1/2 , y = 3/2

Step-by-step explanation:

Solve the following system:

{3 y + x = 5

-2 x - 4 = -5

Hint: | Choose an equation and a variable to solve for.

In the second equation, look to solve for x:

{3 y + x = 5

-2 x - 4 = -5

Hint: | Isolate terms with x to the left hand side.

Add 4 to both sides:

{3 y + x = 5

-2 x = -1

Hint: | Solve for x.

Divide both sides by -2:

{3 y + x = 5

x = 1/2

Hint: | Perform a substitution.

Substitute x = 1/2 into the first equation:

{3 y + 1/2 = 5

x = 1/2

Hint: | Choose an equation and a variable to solve for.

In the first equation, look to solve for y:

{3 y + 1/2 = 5

x = 1/2

Hint: | Isolate terms with y to the left hand side.

Subtract 1/2 from both sides:

{3 y = 9/2

x = 1/2

Hint: | Solve for y.

Divide both sides by 3:

{y = 3/2

x = 1/2

Hint: | Sort results.

Collect results in alphabetical order:

Answer:{x = 1/2 , y = 3/2

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What is the value of x when h(x) = −3?

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Answer:

○ -1

Step-by-step explanation:

Looking closely at this piecewise function, when the line on the left-hand side intersects at -3 = y, x is -1.

i hope this work for you

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2 years ago

Solve for x in the equation 2x^2+3x-7=x^2+5x+39

Shalnov [3]

Hey there, hope I can help!

$\mathrm{Subtract\:}x^2+5x+39\mathrm{\:from\:both\:sides}$
$2x^2+3x-7-\left(x^2+5x+39\right)=x^2+5x+39-\left(x^2+5x+39\right)$

Assuming you know how to simplify this, I will not show the steps but can add them later on upon request
$x^2-2x-46=0$

Lets use the quadratic formula now
$\mathrm{For\:a\:quadratic\:equation\:of\:the\:form\:}ax^2+bx+c=0\mathrm{\:the\:solutions\:are\:}$
$x_{1,\:2}=\frac{-b\pm \sqrt{b^2-4ac}}{2a}$

$\mathrm{For\:} a=1,\:b=-2,\:c=-46: x_{1,\:2}=\frac{-\left(-2\right)\pm \sqrt{\left(-2\right)^2-4\cdot \:1\left(-46\right)}}{2\cdot \:1}$

$\frac{-\left(-2\right)+\sqrt{\left(-2\right)^2-4\cdot \:1\cdot \left(-46\right)}}{2\cdot \:1} \ \textgreater \ \mathrm{Apply\:rule}\:-\left(-a\right)=a \ \textgreater \ \frac{2+\sqrt{\left(-2\right)^2-4\cdot \:1\cdot \left(-46\right)}}{2\cdot \:1}$

Multiply the numbers 2 * 1 = 2
$\frac{2+\sqrt{\left(-2\right)^2-\left(-46\right)\cdot \:1\cdot \:4}}{2}$

$2+\sqrt{\left(-2\right)^2-4\cdot \:1\cdot \left(-46\right)} \ \textgreater \ \sqrt{\left(-2\right)^2-4\cdot \:1\cdot \left(-46\right)}$

$\mathrm{Apply\:rule}\:-\left(-a\right)=a \ \textgreater \ \sqrt{\left(-2\right)^2+1\cdot \:4\cdot \:46} \ \textgreater \ \left(-2\right)^2=2^2, 2^2 = 4$

$\mathrm{Multiply\:the\:numbers:}\:4\cdot \:1\cdot \:46=184 \ \textgreater \ \sqrt{4+184} \ \textgreater \ \sqrt{188} \ \textgreater \ 2 + \sqrt{188}$
$\frac{2+\sqrt{188}}{2} \ \textgreater \ Prime\;factorize\;188 \ \textgreater \ 2^2\cdot \:47 \ \textgreater \ \sqrt{2^2\cdot \:47}$

$\mathrm{Apply\:radical\:rule}: \sqrt[n]{ab}=\sqrt[n]{a}\sqrt[n]{b} \ \textgreater \ \sqrt{47}\sqrt{2^2}$

$\mathrm{Apply\:radical\:rule}: \sqrt[n]{a^n}=a \ \textgreater \ \sqrt{2^2}=2 \ \textgreater \ 2\sqrt{47} \ \textgreater \ \frac{2+2\sqrt{47}}{2}$

$Factor\;2+2\sqrt{47} \ \textgreater \ Rewrite\;as\;1\cdot \:2+2\sqrt{47}$
$\mathrm{Factor\:out\:common\:term\:}2 \ \textgreater \ 2\left(1+\sqrt{47}\right) \ \textgreater \ \frac{2\left(1+\sqrt{47}\right)}{2}$

$\mathrm{Divide\:the\:numbers:}\:\frac{2}{2}=1 \ \textgreater \ 1+\sqrt{47}$

Moving on, I will do the second part excluding the extra details that I had shown previously as from the first portion of the quadratic you can easily see what to do for the second part.

$\frac{-\left(-2\right)-\sqrt{\left(-2\right)^2-4\cdot \:1\cdot \left(-46\right)}}{2\cdot \:1} \ \textgreater \ \mathrm{Apply\:rule}\:-\left(-a\right)=a \ \textgreater \ \frac{2-\sqrt{\left(-2\right)^2-4\cdot \:1\cdot \left(-46\right)}}{2\cdot \:1}$

$\frac{2-\sqrt{\left(-2\right)^2-\left(-46\right)\cdot \:1\cdot \:4}}{2}$

$2-\sqrt{\left(-2\right)^2-4\cdot \:1\cdot \left(-46\right)} \ \textgreater \ 2-\sqrt{188} \ \textgreater \ \frac{2-\sqrt{188}}{2}$

$\sqrt{188} = 2\sqrt{47} \ \textgreater \ \frac{2-2\sqrt{47}}{2}$

$2-2\sqrt{47} \ \textgreater \ 2\left(1-\sqrt{47}\right) \ \textgreater \ \frac{2\left(1-\sqrt{47}\right)}{2} \ \textgreater \ 1-\sqrt{47}$

Therefore our final solutions are
$x=1+\sqrt{47},\:x=1-\sqrt{47}$

Hope this helps!

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2 years ago

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Step-by-step explanation:

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8 0

2 years ago

Will it fit in a pot with a height of 20 cm and a bottom radius of 10 cm of 2.5 liters?

mylen [45]

Answer:

Step-by-step explanation:

Volume of pot = πr²h

= 3.14 * 10 * 10 * 20

= 6280 cubic cm

1000 cubic cm = 1 litre

6280 cubic cm = 6.28 litre

∵ 2.5 litres will fit in the pot as the capacity of pot is 6.28 litre

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3 years ago

Can someone pls help me with these equations

Zielflug [23.3K]

Answer:

1. y = -2x+5

2. y = 3/4 - 3

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3 years ago

Solve by substitutionx+3y=5 -2x-4=-5 ​

Solve by substitutionx+3y=5 -2x-4=-5