Solve Nonlinear Systems By Addition
-$$\begin{array} { l }
x ^ { 2 } - 3 y ^ { 2 } - 1 = 0 \
4 x ^ { 2 } + 3 y ^ { 2 } - 19 = 0
\end{array}$$
A) $\{ ( 2,1 ) , ( 2 , - 1 ) , ( - 2,1 ) , ( - 2 , - 1 ) \}$
B) $\{ ( 1,2 ) , ( - 1,2 ) , ( 1 , - 2 ) , ( - 1 , - 2 ) \}$
C) $\{ ( 2,1 ) , ( - 2 , - 1 ) \}$
D) $\{ ( - 1,2 ) , ( 1 , - 2 ) \}$

Question

Solve Nonlinear Systems By Addition
-$$\begin{array} { l } 
x ^ { 2 } - 3 y ^ { 2 } - 1 = 0 \
4 x ^ { 2 } + 3 y ^ { 2 } - 19 = 0
\end{array}$$
A) $\{ ( 2,1 ) , ( 2 , - 1 ) , ( - 2,1 ) , ( - 2 , - 1 ) \}$
B) $\{ ( 1,2 ) , ( - 1,2 ) , ( 1 , - 2 ) , ( - 1 , - 2 ) \}$
C) $\{ ( 2,1 ) , ( - 2 , - 1 ) \}$
D) $\{ ( - 1,2 ) , ( 1 , - 2 ) \}$

Quizplus · Accepted Answer

Adding the two equations eliminates $y^2$, resulting in $5x^2 - 20 = 0$, which simplifies to $x^2 = 4$, giving $x = \pm 2$. Substituting $x = 2$ into the first equation yields $y^2 = 1$, so $y = \pm 1$. Similarly, substituting $x = -2$ gives the same $y$ values. Thus, the solution set includes $(2, 1)$, $(2, -1)$, $(-2, 1)$, and $(-2, -1)$.

Solve Nonlinear Systems by Addition - x2−3y2−1=04x2+3y2−19=0\begin{array} { l } x ^ { 2 } - 3 y ^ { 2 } - 1 = 0 \\4 x ^ { 2 } + 3 y ^ { 2 } - 19 = 0\end{array}x2−3y2−1=04x2+3y2−19=0​

Solve Nonlinear Systems by Addition
- $\begin{array} { l } x ^ { 2 } - 3 y ^ { 2 } - 1 = 0 \\4 x ^ { 2 } + 3 y ^ { 2 } - 19 = 0\end{array}$