Solve for x and y in the system of equation3x²+4y=17 2x²+5y=2I'll mark brainliest

Mathematics

1 answer:

Novay_Z [31]3 years ago

7 0

Answer:

$x=\pm \sqrt{11} \\y=-4.$

Step-by-step explanation:

$\left \{ {{3x^2+4y=17} \atop {2x^2+5y=2}} \right. \\\\Then\\\\\left \{ {{2 \times (3x^2+4y)=2 \times 17} \atop {3\times(2x^2+5y)=3\times2}} \right. \\\\\left \{ {{6x^2+8y=34 \quad (1)} \atop {6x^2+15y=6\quad (2)}} \right.\\\\(1)-(2) \Rightarrow -7y=28 \Rightarrow y =-4.\\Since \: \: 2x^2+5y=2 \Rightarrow 2x^2=2-5y\\\Rightarrow 2x^2=2+20=22\\\Rightarrow x^2=22:2=11\\\Rightarrow x=\pm \sqrt{11}.$

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Part 4: Use the information provided to write the vertex formula of each parabola.

sergey [27]

Answer: 1. x = (y - 2)² + 8

$\bold{2.\quad x=-\dfrac{1}{2}(y-10)^2}+1$

3. y = 2(x +9)² + 7

<u>Step-by-step explanation:</u>

Notes: Vertex form is: y =a(x - h)² + k or x =a(y - k)² + h

(h, k) is the vertex
point of vertex is midpoint of focus and directrix: $\dfrac{focus+directrix}{2}$

$\bullet\quad a=\dfrac{1}{4p}$

p is the distance from the vertex to the focus

$focus = \bigg(\dfrac{-31}{4},2\bigg)\qquad directrix: x=\dfrac{-33}{4}\\\\\text{Since directrix is x, then the x-value of the vertex is:}\\\\\dfrac{focus+directrix}{2}=\dfrac{\frac{-31}{4}+\frac{-33}{4}}{2}=\dfrac{\frac{-64}{4}}{2}=\dfrac{-16}{2}=-8\\\\\text{The y-value of the vertex is given by the focus as: 2}\\\\\text{vertex (h, k)}=(-8,2)$

Now let's find the a-value:

$p=focus-vertex\\\\p=\dfrac{-31}{4}-\dfrac{-32}{4}=\dfrac{1}{4}\\\\\\a=\dfrac{1}{4p}=\dfrac{1}{4(\frac{1}{4})}=\dfrac{1}{1}=1$

Now, plug in a = 1 and (h, k) = (-8, 2) into the equation x =a(y - k)² + h

x = (y - 2)² + 8

***************************************************************************************

$focus = \bigg(\dfrac{1}{2},10\bigg)\qquad directrix: x=\dfrac{3}{2}\\\\\text{Since directrix is x, then the x-value of the vertex is:}\\\\\dfrac{focus+directrix}{2}=\dfrac{\frac{1}{2}+\frac{3}{2}}{2}=\dfrac{\frac{4}{2}}{2}=\dfrac{2}{2}=1\\\\\text{The y-value of the vertex is given by the focus as: 10}\\\\\text{vertex (h, k)}=(1,10)$

Now let's find the a-value:

$p=focus-vertex\\\\p=\dfrac{1}{2}-\dfrac{2}{2}=\dfrac{-1}{2}\\\\\\a=\dfrac{1}{4p}=\dfrac{1}{4(\frac{-1}{2})}=\dfrac{1}{-2}=-\dfrac{1}{2}$

Now, plug in a = -1/2 and (h, k) = (1, 10) into the equation x =a(y - k)² + h

$\bold{x=-\dfrac{1}{2}(y-10)^2}+1$

***************************************************************************************

$focus = \bigg(-9,\dfrac{57}{8}\bigg)\qquad directrix: y=\dfrac{55}{8}\\\\\text{Since directrix is y, then the y-value of the vertex is:}\\\\\dfrac{focus+directrix}{2}=\dfrac{\frac{57}{8}+\frac{55}{8}}{2}=\dfrac{\frac{112}{8}}{2}=\dfrac{14}{2}=7\\\\\text{The x-value of the vertex is given by the focus as: -9}\\\\\text{vertex (h, k)}=(-9,7)$