Solve the system of equations using elimination.
-$$\left\{ \begin{array} { l }
2 x ^ { 2 } + y ^ { 2 } = 17 \
3 x ^ { 2 } - 2 y ^ { 2 } = - 6
\end{array} \right.$$
A) $x = 2 , y = 3 ; x = 2 , y = - 3 ; x = - 2 , y = 3 ; x = - 2 , y = - 3$
B) $x = 2 , y = - 3 ; x = - 2 , y = 3$
C) $x = 1 , y = 3 ; x = - 1 , y = - 3$
D) $\mathrm { x } = 1 , \mathrm { y } = 3 ; \mathrm { x } = 1 , \mathrm { y } = - 3 ; \mathrm { x } = - 1 , \mathrm { y } = 3 ; \mathrm { x } = - 1 , \mathrm { y } = - 3$

Question

Solve the system of equations using elimination.
-$$\left\{ \begin{array} { l } 
2 x ^ { 2 } + y ^ { 2 } = 17 \
3 x ^ { 2 } - 2 y ^ { 2 } = - 6
\end{array} \right.$$
A) $x = 2 , y = 3 ; x = 2 , y = - 3 ; x = - 2 , y = 3 ; x = - 2 , y = - 3$
B) $x = 2 , y = - 3 ; x = - 2 , y = 3$
C) $x = 1 , y = 3 ; x = - 1 , y = - 3$
D) $\mathrm { x } = 1 , \mathrm { y } = 3 ; \mathrm { x } = 1 , \mathrm { y } = - 3 ; \mathrm { x } = - 1 , \mathrm { y } = 3 ; \mathrm { x } = - 1 , \mathrm { y } = - 3$

Quizplus · Accepted Answer

To solve the system of equations using elimination, we can multiply the first equation by 2 and the second equation by 1 to make the coefficients of $y^2$ opposites:$$4x^2 + 2y^2 = 34$$$$3x^2 - 2y^2 = -6$$Adding these equations eliminates $y^2$, resulting in:$$7x^2 = 28$$$$x^2 = 4$$$$x = \pm 2$$Substituting $x = 2$ into the first equation:$$2(2)^2 + y^2 = 17$$$$8 + y^2 = 17$$$$y^2 = 9$$$$y = \pm 3$$Thus, the solutions are $x = 2, y = 3; x = 2, y = -3; x = -2, y = 3; x = -2, y = -3$.

Solve the System of Equations Using Elimination {2x2+y2=173x2−2y2=−6\left\{ \begin{array} { l } 2 x ^ { 2 } + y ^ { 2 } = 17 \\3 x ^ { 2 } - 2 y ^ { 2 } = - 6\end{array} \right.{2x2+y2=173x2−2y2=−6​

Solve the System of Equations Using Elimination $\left\{ \begin{array} { l } 2 x ^ { 2 } + y ^ { 2 } = 17 \\3 x ^ { 2 } - 2 y ^ { 2 } = - 6\end{array} \right.$