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# Pls help

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Another question for me to ask again.

1.Given the product of 2x+3y and 3y+2x is z. Find the value of y when x=3.2 and z=457.96

2.3x+5y=-2

-2x-3y=2

3.sqrt3(250/1024)

Pls help

twhtan  Sep 21, 2017
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#1
+85829
+1

Note that we really have that

(2x + 3y)^2 = z   ...so....

(2(3.2)  + 3y)^2  = 457.96       simplify inside the parentheses

( 6.4 + 3y)^2  = 457.96          take the positive and negative square roots

6.4 + 3y   = ±√457.96            subtract 6.4 from both sides

3y  = ±√457.96 - 6.4             divide both sides by 3

y  = [ ±√457.96 - 6.4 ] / 3

So we have that   y  = [√457.96 - 6.4 ] / 3   = 5

And        y  = [ - √457.96 - 6.4 ] / 3   ≈  -9.2667

2.    3x+5y= -2   multiply through by 3     →  9x + 15y  =  - 6        (1)

-2x-3y=  2    multiply through by 5    → -10x - 15y  =  10        (2)

Add (1) and (2)   and we get that

-1x  = 4      divide both sides by -1

x  = -4

Subbing this back into any of the equations for x to find y, we have

3 (-4)  + 5y  = -2

-12  + 5y  = -2        add 12 to both sides

5y  = 10                divide both sides  by 5

y  = 2

3. sqrt3(250/1024)

I'm guessing that this might actually be   ∛  [ 250 / 1024]  ....if so, we have....

∛ [ (2 * 125)  / ( 2 / 512) ] =

∛ [  (2 * 5^3)  / ( 2 * 8^3) ]   =

(5/8) ∛ (2/2)  =

(5/8)∛ 1  =

(5/8)

CPhill  Sep 21, 2017
#2
+1873
+1

1)

If $$(2x+3y)(3y+2x)=z$$, according to the given information, and $$x=3.2$$ and $$z=457.96$$, just plug those values in to solve for y:

$$(2x+3y)(3y+2x)=z$$ Plug in the appropriate values for the given variables of x and z.
$$(2*3.2+3y)(2*3.2+3y)=457.96$$ Simplify what is inside the parentheses first.
$$(6.4+3y)(6.4+3y)=457.96$$ You might notice that both the multiplicand and multiplier are the same, which means that we can make this equation a tad simpler.
$$(6.4+3y)^2=457.96$$ Take the square root of both sides. Of course, this breaks the equation up into its positive and negative answer.
$$6.4+3y=\pm\sqrt{457.96}$$ Although it may not be obvious, the square root of happens to work out nicely.
$$6.4+3y=\pm21.4$$ To solve for y, we must break up the equation.
 $$6.4+3y=21.4$$ $$6.4+3y=-21.4$$

Now, subtract by 6.4 in both equations.
 $$3y=15$$ $$3y=-27.8$$

Divide by 3 on both sides.
 $$y_1=5$$ $$y_2=-\frac{27.8}{3}*\frac{10}{10}=-\frac{278}{30}=-9.2\overline{6}$$

Both of these y-values satisfy the equation, and these are the solutions.

2)

This is a system of equations. I usually refrain from using the elimination method here because it is difficult to showcase. Therefore, I will use the substitution method.

I will solve for y in equation 2:

 $$-2x-3y=2$$ Add 2x to both sides. $$-3y=2x+2$$ Divide by -3 to isolate y. $$y=-\frac{2x+2}{3}$$

Plug this value for y into equation 1 and then solve for x.

 $$3x+5y=-2$$ Plug in the value for y that was determined from the previous equation. $$3x+5*\frac{2x+2}{-3}=-2$$ Do the multiplication first to simplify this monstrosity. $$5*\frac{2x+2}{-3}=\frac{5(2x+2)}{-3}=\frac{10x+10}{-3}$$ Now, reinsert this back into the original equation. $$3x+\frac{10x+10}{-3}=-2$$ Multiply by -3 on all sides to get rid of the fraction. $$-9x+10x+10=6$$ Combine the like terms on the left hand side. $$x+10=6$$ Subtract 10 on both sides. $$x=-4$$

Now, plug x=-4 into either equation and solve for y. I'll choose equation 2 because it look easier to do:

 $$-2x-3y=2$$ Substitute all x's for -4. $$-2*-4-3y=2$$ $$8-3y=2$$ Subtract by 8 on both sides. $$-3y=-6$$ Divide by -3 to isolate y. $$y=2$$

Therefore, the coordinate where both lines intersect is $$(-4,2)$$.

TheXSquaredFactor  Sep 21, 2017

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