tatiana7121
Answered

To solve the system of equations below, Zach isolated [tex]$x^2$[/tex] in the first equation and then substituted it into the second equation. What was the resulting equation?

[tex]\[
\left\{
\begin{array}{l}
x^2 + y^2 = 25 \\
\frac{x^2}{16} - \frac{y^2}{9} = 1
\end{array}
\right.
\][/tex]

A. [tex]\(\frac{y^2 - 25}{16} - \frac{y^2}{9} = 1\)[/tex]

B. [tex]\(\frac{25 - y^2}{16} - \frac{y^2}{9} = 1\)[/tex]

C. [tex]\(\frac{x^2}{16} - \frac{25 - y^2}{9} = 1\)[/tex]

D. [tex]\(\frac{x^2}{16} - \frac{y^2 - 25}{9} = 1\)[/tex]



Answer :

GinnyAnswer
To solve the given system of equations:

[tex]\[ \left\{\begin{array}{l} x^2 + y^2 = 25 \\ \frac{x^2}{16} - \frac{y^2}{9} = 1 \end{array}\right. \][/tex]

we can isolate [tex]\(x^2\)[/tex] from the first equation and then substitute it into the second equation.

1. Start with the first equation:

[tex]\[ x^2 + y^2 = 25 \][/tex]

Isolate [tex]\(x^2\)[/tex] by subtracting [tex]\(y^2\)[/tex] from both sides:

[tex]\[ x^2 = 25 - y^2 \][/tex]

2. Substitute this expression for [tex]\(x^2\)[/tex] into the second equation:

[tex]\[ \frac{x^2}{16} - \frac{y^2}{9} = 1 \][/tex]

Substitute [tex]\(x^2 = 25 - y^2\)[/tex]:

[tex]\[ \frac{25 - y^2}{16} - \frac{y^2}{9} = 1 \][/tex]

Thus, the resulting equation is:

[tex]\[ \frac{25 - y^2}{16} - \frac{y^2}{9} = 1 \][/tex]

This corresponds to option B.

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