Question

Which of these expressions equal 15 when x = and y=3? Circle

Answer 1

Answer:

Option 1: 4(2y - 4x) -1

Step-by-step explanation:

Hello!

Given:

• x = 0.5
• y = 3

Plug in the values for x and y into each equation and see which one outputs 15.

4(2y - 4x) -1

• 4(2(3) - 4(0.5)) - 1
• 4(6 - 2) - 1
• 4(4) - 1
• 16-1
• 15

(x² + 1) + 2x + 3y

• (0.5² + 1) + 2(0.5) + 3(3)
• 1.25 + 1 + 9
• 11.25

4x²+2y³-10

• 4(0.5²) + 2(3³) - 10
• 4(0.25) + 2(27) - 10
• 1 + 54 - 10
• 55 - 10
• 45

xy+3+20x

• (0.5)(3) + 3 + 20(0.5)
• 1.5 + 3 + 10
• 4.5 + 10
• 14.5

The only option that works is the first option.

Answer 2

Answer:

$4 (2y-4x)-1$

Step-by-step explanation:

Given:

• $x =\dfrac{1}{2}$

• $y=3$

To find which expressions equal 15, substitute the given values of x and y into the expressions and evaluate:

$\begin{aligned}4(2y-4x)-1 &= 4 \left(2(3)-4 \left(\dfrac{1}{2}\right)\right)-1\\& = 4 \left(6-2\right)-1\\& = 4 \left(4\right)-1\\& = 16-1\\& = 15\\\end{aligned}$

$\begin{aligned}(x^2+1)+2x+3y & = \left(\left(\dfrac{1}{2}\right)^2+1\right)+2 \left(\dfrac{1}{2}\right)+3(3)\\& = \left(\left\dfrac{1}{4}+1\right)+1+9\\& = \dfrac{5}{4}+1+1+9\\& = \dfrac{45}{4}\end{aligned}$

$\begin{aligned}4x^2+2y^3-10 & = 4\left(\dfrac{1}{2}\right)^2+2(3)^3-10\\& = 4\left(\dfrac{1}{4}\right)+2(27)-10\\& = 1+54-10\\& = 45\end{aligned}$

$\begin{aligned}xy+3+20x & = \left(\dfrac{1}{2}\right)(3)+3+20\left(\dfrac{1}{2}\right)\\& = \dfrac{3}{2}+3+10\\& = \dfrac{29}{2}\end{aligned}$

Therefore, the only expression that equals 15 is:

  $4(2y-4x)-1$