Asked by Annalee Geels on Mar 10, 2024

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Graph the equations to determine whether the system has any solutions. Find any solutions that exist. {2x+y=12x2+y2=36\left\{ \begin{array} { r } 2 x + y = 12 \\x ^ { 2 } + y ^ { 2 } = 36\end{array} \right.{2x+y=12x2+y2=36

A) (6,0) (−185,−245) ( 6,0 ) \left( - \frac { 18 } { 5 } , - \frac { 24 } { 5 } \right) (6,0) (518,524)
B) (−6,0) (185,245) ( - 6,0 ) \left( \frac { 18 } { 5 } , \frac { 24 } { 5 } \right) (6,0) (518,524)
C) (6,0) (185,245) ( 6,0 ) \left( \frac { 18 } { 5 } , \frac { 24 } { 5 } \right) (6,0) (518,524)
D) (−6,0) (−185,−245) ( - 6,0 ) \left( - \frac { 18 } { 5 } , - \frac { 24 } { 5 } \right) (6,0) (518,524)
E) no solution exists

Graph Equations

The process of plotting solutions to an equation on a coordinate plane to visualize the relationship between variables.

System of Equations

A set of two or more equations with the same variables, where the solution is the set of values that satisfy all equations simultaneously.

  • Construct graphs of equations and determine whether solutions are present.
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Verified Answer

DS
Dariell Simmons

Mar 10, 2024

Final Answer :
C
Explanation :
We can solve for $y$ in the first equation to get $y=12-2x$ and substitute into the second equation: x2+(12−2x)2=36  ⟹  5x2−48x+72=0x^2+(12-2x)^2=36 \implies 5x^2-48x+72=0x2+(122x)2=365x248x+72=0 Solving for $x$ using the quadratic formula we get: x=48±482−4(5)(72)10=24±6215x=\frac{48 \pm \sqrt{48^2-4(5)(72)}}{10}= \frac{24 \pm 6\sqrt{21}}{5}x=1048±4824(5)(72)=524±621 Substituting $x$ back into the first equation we get the corresponding solutions for $y$: x=24+6215  ⟹  y=24−2(24+621)5=−245−12215x=\frac{24 + 6\sqrt{21}}{5} \implies y=\frac{24 - 2(24 + 6\sqrt{21})}{5}=-\frac{24}{5} -\frac{12\sqrt{21}}{5}x=524+621y=5242(24+621)=52451221 x=24−6215  ⟹  y=24−2(24−621)5=−245+12215x=\frac{24 - 6\sqrt{21}}{5} \implies y=\frac{24 - 2(24 - 6\sqrt{21})}{5}=-\frac{24}{5} +\frac{12\sqrt{21}}{5}x=524621y=5242(24621)=524+51221 Therefore, the system has two solutions: $\left(\frac{24 + 6\sqrt{21}}{5},-\frac{24}{5} -\frac{12\sqrt{21}}{5}\right)$ and $\left(\frac{24 - 6\sqrt{21}}{5},-\frac{24}{5} +\frac{12\sqrt{21}}{5}\right)$. The correct choice is $\boxed{\textbf{(C)}}$.