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2 years ago

Exploring Systems of Linear Equations 2x +3y =8 and 3x+y= -2

Mathematics

1 answer:

butalik [34]2 years ago

3 0

The solution of the linear equations will be ( -2, 4).

<h3>What is an equation?</h3>

It is defined as the relation between two variables, if we plot the graph of the linear equation we will get a straight line.

Given equations are:-

2x +3y = 8 and 3x+y= -2

Solving the equations by elimination method:-

2x +3y = 8

3x+y= -2

Multiply the second equation by 3 and subtract from the first equation.

2x +3y = 8

-9x -3y = 6

----------------

-7x = 14

x = -2

Out of the value of x in any one equation, we will get the value of y.

3x+y= -2

3 ( -2) + y = -2

-6 + y = -2

y = 4

The graph of the equations is also attached with the answer below.

Therefore the solution of the linear equations will be ( -2, 4).

The complete question is given below:-

Exploring Systems of Linear Equations 2x +3y =8 and 3x+y= -2. Find the value of x and y and draw a graph for the system of linear equations.

To know more about equations follow

brainly.com/question/2972832

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Solve for x.

Ivahew [28]

ANSWER

$x=\frac{2-\sqrt{10}} {3}$

or

$x=\frac{\sqrt{10}+2} {3}$

We have

$3x^2-4x-2=0$

Since we cannot factor easily, we complete the square.

Adding 2 to both sides give,

$3x^2-4x=2$

Dividing through by 3 gives

$x^2-\frac{4}{3}x= \frac{2}{3}$

Adding $(-\frac{2}{3})^2$ to both sides gives

$x^2-\frac{4}{3}x+(-\frac{2}{3})^2= \frac{2}{3}+(-\frac{2}{3})^2$

The expression on the Left Hand side is a perfect square.

$(x-\frac{2}{3})^2= \frac{2}{3}+\frac{4}{9}$

$\Rightarrow (x-\frac{2}{3})^2= \frac{10}{9}$

$\Rightarrow (x-\frac{2}{3})=\pm \sqrt{\frac{10}{9}}$

$\Rightarrow (x)=\frac{2}{3} \pm {\frac{\sqrt{10}}{3}$

Splitting the plus or minus sign gives

$x=\frac{2- \sqrt{10}} {3}$

or

$x=\frac{\sqrt{10}+2} {3}$

3 0

3 years ago

Read 2 more answers

Please help me for the love of God if i fail I have to repeat the class

Elena-2011 [213]

$\theta$ is in quadrant I, so $\cos\theta>0$ .

$x$ is in quadrant II, so $\sin x>0$ .

Recall that for any angle $\alpha$ ,

$\sin^2\alpha+\cos^2\alpha=1$

Then with the conditions determined above, we get

$\cos\theta=\sqrt{1-\left(\dfrac45\right)^2}=\dfrac35$

and

$\sin x=\sqrt{1-\left(-\dfrac5{13}\right)^2}=\dfrac{12}{13}$

Now recall the compound angle formulas:

$\sin(\alpha\pm\beta)=\sin\alpha\cos\beta\pm\cos\alpha\sin\beta$

$\cos(\alpha\pm\beta)=\cos\alpha\cos\beta\mp\sin\alpha\sin\beta$

$\sin2\alpha=2\sin\alpha\cos\alpha$

$\cos2\alpha=\cos^2\alpha-\sin^2\alpha$

as well as the definition of tangent:

$\tan\alpha=\dfrac{\sin\alpha}{\cos\alpha}$

Then

1. $\sin(\theta+x)=\sin\theta\cos x+\cos\theta\sin x=\dfrac{16}{65}$

2. $\cos(\theta-x)=\cos\theta\cos x+\sin\theta\sin x=\dfrac{33}{65}$

3. $\tan(\theta+x)=\dfrac{\sin(\theta+x)}{\cos(\theta+x)}=-\dfrac{16}{63}$

4. $\sin2\theta=2\sin\theta\cos\theta=\dfrac{24}{25}$

5. $\cos2x=\cos^2x-\sin^2x=-\dfrac{119}{169}$

6. $\tan2\theta=\dfrac{\sin2\theta}{\cos2\theta}=-\dfrac{24}7$

7. A bit more work required here. Recall the half-angle identities:

$\cos^2\dfrac\alpha2=\dfrac{1+\cos\alpha}2$

$\sin^2\dfrac\alpha2=\dfrac{1-\cos\alpha}2$

$\implies\tan^2\dfrac\alpha2=\dfrac{1-\cos\alpha}{1+\cos\alpha}$

Because $x$ is in quadrant II, we know that $\dfrac x2$ is in quadrant I. Specifically, we know $\dfrac\pi2$ , so $\dfrac\pi4$ . In this quadrant, we have $\tan\dfrac x2>0$ , so