Answer :
To determine which system is equivalent to the given one, we can multiply the equations in the original system to match the options provided. The original system is:
1) [tex]\(5x^2 + 6y^2 = 50\)[/tex]
2) [tex]\(7x^2 + 2y^2 = 10\)[/tex]
We need to find an equivalent system by manipulating the original equations. Let's explore the options:
Option 1:
[tex]\[
\begin{aligned}
&5x^2 + 6y^2 = 50 \quad \text{(same as equation 1)} \\
&-21x^2 - 6y^2 = 10
\end{aligned}
\][/tex]
This doesn't derive from a simple multiplication of either original equation, so it's not equivalent.
Option 2:
[tex]\[
\begin{aligned}
&5x^2 + 6y^2 = 50 \quad \text{(same as equation 1)} \\
&-21x^2 - 6y^2 = 30
\end{aligned}
\][/tex]
Again, this doesn't derive from a straightforward multiplication or transformation of the second original equation, so it's not equivalent.
Option 3:
[tex]\[
\begin{aligned}
&35x^2 + 42y^2 = 250 \\
&-35x^2 - 10y^2 = -50
\end{aligned}
\][/tex]
Let's transform the original equations to see if they can match this option:
- Multiply the first original equation [tex]\(5x^2 + 6y^2 = 50\)[/tex] by 7:
[tex]\[
7 \times (5x^2 + 6y^2) = 7 \times 50 \quad \Rightarrow \quad 35x^2 + 42y^2 = 350
\][/tex]
- Multiply the second original equation [tex]\(7x^2 + 2y^2 = 10\)[/tex] by -5:
[tex]\[
-5 \times (7x^2 + 2y^2) = -5 \times 10 \quad \Rightarrow \quad -35x^2 - 10y^2 = -50
\][/tex]
However, note that when we correctly do this step, it fits better with option 4 in terms of the results displayed.
Option 4:
[tex]\[
\begin{aligned}
&35x^2 + 42y^2 = 350 \\
&-35x^2 - 10y^2 = -50
\end{aligned}
\][/tex]
These transformations from the original equations correctly match with this set of equations. Therefore, this option is equivalent to the original system.
Thus, after considering each option, the one that truly represents an equivalent system by a simple multiplication of the original equations is Option 4.
1) [tex]\(5x^2 + 6y^2 = 50\)[/tex]
2) [tex]\(7x^2 + 2y^2 = 10\)[/tex]
We need to find an equivalent system by manipulating the original equations. Let's explore the options:
Option 1:
[tex]\[
\begin{aligned}
&5x^2 + 6y^2 = 50 \quad \text{(same as equation 1)} \\
&-21x^2 - 6y^2 = 10
\end{aligned}
\][/tex]
This doesn't derive from a simple multiplication of either original equation, so it's not equivalent.
Option 2:
[tex]\[
\begin{aligned}
&5x^2 + 6y^2 = 50 \quad \text{(same as equation 1)} \\
&-21x^2 - 6y^2 = 30
\end{aligned}
\][/tex]
Again, this doesn't derive from a straightforward multiplication or transformation of the second original equation, so it's not equivalent.
Option 3:
[tex]\[
\begin{aligned}
&35x^2 + 42y^2 = 250 \\
&-35x^2 - 10y^2 = -50
\end{aligned}
\][/tex]
Let's transform the original equations to see if they can match this option:
- Multiply the first original equation [tex]\(5x^2 + 6y^2 = 50\)[/tex] by 7:
[tex]\[
7 \times (5x^2 + 6y^2) = 7 \times 50 \quad \Rightarrow \quad 35x^2 + 42y^2 = 350
\][/tex]
- Multiply the second original equation [tex]\(7x^2 + 2y^2 = 10\)[/tex] by -5:
[tex]\[
-5 \times (7x^2 + 2y^2) = -5 \times 10 \quad \Rightarrow \quad -35x^2 - 10y^2 = -50
\][/tex]
However, note that when we correctly do this step, it fits better with option 4 in terms of the results displayed.
Option 4:
[tex]\[
\begin{aligned}
&35x^2 + 42y^2 = 350 \\
&-35x^2 - 10y^2 = -50
\end{aligned}
\][/tex]
These transformations from the original equations correctly match with this set of equations. Therefore, this option is equivalent to the original system.
Thus, after considering each option, the one that truly represents an equivalent system by a simple multiplication of the original equations is Option 4.
