+9 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
+130091
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)
+2446 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. | ||
| Now, subtract by 6.4 in both equations. | ||
| Divide by 3 on both sides. | ||
| 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 \((-4,2)\).
