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nordsb [41]
2 years ago
14

Solve the system of equations using the elimination method. 4x+3y=16 x=-3y+13

Mathematics
2 answers:
Ipatiy [6.2K]2 years ago
7 0
Step by Step Solution
More Icon
System of Linear Equations entered :

[1] -4x - 3y = -7
[2] 16x - 3y = 13
Graphic Representation of the Equations :

-3y - 4x = -7 -3y + 16x = 13



Solve by Substitution :

// Solve equation [2] for the variable x


[2] 16x = 3y + 13

[2] x = 3y/16 + 13/16
// Plug this in for variable x in equation [1]

[1] -4•(3y/16+13/16) - 3y = -7
[1] - 15y/4 = -15/4
[1] - 15y = -15
// Solve equation [1] for the variable y


[1] 15y = 15

[1] y = 1
// By now we know this much :

x = 3y/16+13/16
y = 1
// Use the y value to solve for x

x = (3/16)(1)+13/16 = 1
Solution :
{x,y} = {1,1}
Gemiola [76]2 years ago
3 0

Answer:

 x=1   y=4

Step-by-step explanation:

1.  4x+3y=16

   x=-3y+13

2. 4x+3y=16

    x+3y=13

3.  4x+3y=16

     -x-3y=-13

4.  3x=3

5.   x=1

6. (1)=-3y+13

7.  1-13=-3y

8.   -12=-3y

9.    y=4

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<img src="https://tex.z-dn.net/?f=prove%20that%5C%20%20%5Ctextless%20%5C%20br%20%2F%5C%20%20%5Ctextgreater%20%5C%20%5Cfrac%20%7B
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\large \bigstar \frak{ } \large\underline{\sf{Solution-}}

Consider, LHS

\begin{gathered}\rm \: \dfrac { \tan \theta + \sec \theta - 1 } { \tan \theta - \sec \theta + 1 } \\ \end{gathered}

We know,

\begin{gathered}\boxed{\sf{  \:\rm \: {sec}^{2}x - {tan}^{2}x = 1 \: \: }} \\ \end{gathered}  \\  \\  \text{So, using this identity, we get} \\  \\ \begin{gathered}\rm \: = \:\dfrac { \tan \theta + \sec \theta - ( {sec}^{2}\theta - {tan}^{2}\theta )} { \tan \theta - \sec \theta + 1 } \\ \end{gathered}

We know,

\begin{gathered}\boxed{\sf{  \:\rm \: {x}^{2} - {y}^{2} = (x + y)(x - y) \: \: }} \\ \end{gathered}  \\

So, using this identity, we get

\begin{gathered}\rm \: = \:\dfrac { \tan \theta + \sec \theta - (sec\theta + tan\theta )(sec\theta - tan\theta )} { \tan \theta - \sec \theta + 1 } \\ \end{gathered}

can be rewritten as

\begin{gathered}\rm\:=\:\dfrac {(\sec \theta + tan\theta ) - (sec\theta + tan\theta )(sec\theta -tan\theta )} { \tan \theta - \sec \theta + 1 } \\ \end{gathered} \\  \\  \\\begin{gathered}\rm \: = \:\dfrac {(\sec \theta + tan\theta ) \: \cancel{(1 - sec\theta + tan\theta )}} { \cancel{ \tan \theta - \sec \theta + 1} } \\ \end{gathered} \\  \\  \\\begin{gathered}\rm \: = \:sec\theta + tan\theta \\\end{gathered} \\  \\  \\\begin{gathered}\rm \: = \:\dfrac{1}{cos\theta } + \dfrac{sin\theta }{cos\theta } \\ \end{gathered} \\  \\  \\\begin{gathered}\rm \: = \:\dfrac{1 + sin\theta }{cos\theta } \\ \end{gathered}

<h2>Hence,</h2>

\begin{gathered} \\ \rm\implies \:\boxed{\sf{  \:\rm \: \dfrac { \tan \theta + \sec \theta - 1 } { \tan \theta - \sec \theta + 1 } = \:\dfrac{1 + sin\theta }{cos\theta } \: \: }} \\ \\ \end{gathered}

\rule{190pt}{2pt}

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