Fiona wrote the linear equation y = y equals StartFraction 2 over 5 EndFraction x minus 5.x – 5. When Henry wrote his equation, they discovered that his equation had all the same solutions as Fiona’s. Which equation could be Henry’s? x – x minus StartFraction 5 over 4 EndFraction y equals StartFraction 25 over 4 EndFraction.y = x – x minus StartFraction 5 over 2 EndFraction y equals StartFraction 25 over 4 EndFraction.y = x – x minus StartFraction 5 over 4 EndFraction y equals StartFraction 25 over 2 EndFraction.y = x – x minus StartFraction 5 over 2 EndFraction y equals StartFraction 25 over 2 EndFraction.y =

Respuesta :

Answer:

D. [tex]x-\frac{5}{2}y = \frac{25}{2}[/tex]

Step-by-step explanation:

Given

[tex]y = \frac{2}{5}x - 5[/tex]

Required

Determine its equivalent

From the list of given options, the correct answer is

[tex]x - \frac{5}{2}y = \frac{25}{2}[/tex]

This is shown as follows;

[tex]y = \frac{2}{5}x - 5[/tex]

Multiply both sides by [tex]\frac{5}{2}[/tex]

[tex]\frac{5}{2} * y = \frac{5}{2} * (\frac{2}{5}x - 5)[/tex]

Open Bracket

[tex]\frac{5}{2} * y = \frac{5}{2} * \frac{2}{5}x - \frac{5}{2} *5[/tex]

[tex]\frac{5}{2}y = x - \frac{25}{2}[/tex]

Subtract x from both sides

[tex]\frac{5}{2}y - x = x -x - \frac{25}{2}[/tex]

[tex]\frac{5}{2}y - x = - \frac{25}{2}[/tex]

Multiply both sides by -1

[tex]-1 * \frac{5}{2}y - x * -1 = - \frac{25}{2} * -1[/tex]

[tex]-\frac{5}{2}y + x = \frac{25}{2}[/tex]

Reorder

[tex]x-\frac{5}{2}y = \frac{25}{2}[/tex]

Hence, the correct option is D

[tex]x-\frac{5}{2}y = \frac{25}{2}[/tex]

Answer:

The 4th option

Step-by-step explanation: