Mathematics: Find the solution of the system of equations. 2x + 3y = -2 4x + 9y = 14

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Find the solution of the system of equations. - id211996228, vanitycarraway2000

Ответ:

ophanteen

17.04.2021

Mathematics

x = -10

y = 6

Step-by-step explanation:

here's the solution : -

=》2x + 3y = -2

=》4x + 9y = 14

multiplying the equation 1 by 2

=》2( 2x + 3x = -2 )

=》4x + 6y = -4

now, subtracting the equation 1 from equation 2

=》4x + 9y - 4x - 6y = 14 - (-4)

=》3y = 18

=》y = 18 ÷ 3

=》y = 6

now, plugging the value of y as 6 in equation 2, we get : -

=》4x + 9y = 14

=》4x + (9 × 6) = 14

=》4x + 54 = 14

=》4x = 14 - 54

=》x = -40 ÷ 4

=》x = -10

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Ответ:

queenbb3787

20.11.2022

Mathematics

$\large\begin{array}{l} \textsf{Solve the trigonometric equation}\\\\ \mathsf{\sqrt{3}\,sin\,x+cos\,x=\sqrt{3}}\qquad\quad\mathsf{0^\circ$

$\large\begin{array}{l} \textsf{Multiply both sides by }\mathsf{\dfrac{1}{2}:}\\\\ \mathsf{\dfrac{1}{2}\cdot \left(\sqrt{3}\,sin\,x+cos\,x\right)=\dfrac{1}{2}\cdot \sqrt{3}}\\\\ \mathsf{\dfrac{\sqrt{3}}{2}\,sin\,x+\dfrac{1}{2}\,cos\,x=\dfrac{\sqrt{3}}{2}\qquad\quad(i)} \end{array}$

$\large\begin{array}{l} \textsf{But,}\\\\ \left\{\! \begin{array}{rcl} \mathsf{\dfrac{\sqrt{3}}{2}}&\!\!=\!\!&\mathsf{cos\,30^\circ}\\\\ \mathsf{\dfrac{1}{2}}&\!\!=\!\!&\mathsf{sin\,30^\circ} \end{array} \right. \end{array}$

$\large\begin{array}{l} \textsf{so (i) becomes}\\\\ \mathsf{cos\,30^\circ\,sin\,x+sin\,30^\circ\,cos\,x=\dfrac{\sqrt{3}}{2}}\\\\ \mathsf{sin\,x\,cos\,30^\circ+cos\,x\,sin\,30^\circ=\dfrac{\sqrt{3}}{2}\qquad\quad(ii)} \end{array}$

$\large\begin{array}{l} \textsf{Notice that the left side of (ii) is the expanded form of the}\\\textsf{sine of a sum:}\\\\ \mathsf{sin(\alpha+\beta)=sin\,\alpha\,cos\,\beta+cos\,\alpha\,sin\,\beta}\\\\ \textsf{where }\mathsf{\alpha=x~~and~~\beta=30^\circ.} \end{array}$

$\large\begin{array}{l} \textsf{So we can use this to rewrite (ii) as}\\\\ \mathsf{sin(x+30^\circ)=\dfrac{\sqrt{3}}{2}}\qquad\quad\Big(\textsf{but }\mathsf{\dfrac{\sqrt{3}}{2}=sin\,60^\circ}\Big)\\\\ \mathsf{sin(x+30^\circ)=sin\,60^\circ\qquad\quad(iii)} \end{array}$

$\large\begin{array}{l} \textsf{From the equality of sines above, we get}\\\\ \begin{array}{rcl} \mathsf{x+30^\circ=60^\circ+k\cdot 360^\circ}&~\textsf{ or }~&\mathsf{x+30^\circ=(180^\circ-60^\circ)+k\cdot 360^\circ}\\\\ \mathsf{x+30^\circ=60^\circ+k\cdot 360^\circ}&~\textsf{ or }~&\mathsf{x+30^\circ=120^\circ+k\cdot 360^\circ}\\\\ \mathsf{x=60^\circ-30^\circ+k\cdot 360^\circ}&~\textsf{ or }~&\mathsf{x=120^\circ-30^\circ+k\cdot 360^\circ} \end{array} \end{array}$

$\large\begin{array}{l} \boxed{\begin{array}{rcl} \mathsf{x=30^\circ+k\cdot 360^\circ}&~\textsf{ or }~&\mathsf{x=90^\circ+k\cdot 360^\circ} \end{array}} \\\\\\\textsf{where k is an integer.}\end{array}$

________

$\large\begin{array}{l} \textsf{But for this question, we have also to restrict the solutions}\\\textsf{to this specific interval: }\mathsf{0^\circ$

$\large\begin{array}{l} \bullet~~\textsf{For }\mathsf{k=0,}\\\\ \boxed{\begin{array}{rcl} \mathsf{x=30^\circ}&~\textsf{ or }~&\mathsf{x=90^\circ} \end{array}}\\\\\uparrow\\\\ \textsf{those are the solutions for the equation.}\\\\\\ \textsf{Solution set: }\mathsf{S=\{30^\circ,\,90^\circ\}} \end{array}$

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$\large\textsf{I hope this helps. :-)}$

Tags: trigonometric equation trig sine sin cosine cos sum identity arc angle solve solution

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Find the solution of the system of equations.
2x + 3y = -2
4x + 9y = 14

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