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4 years ago

7x+12=2(1-x)-x solve

Mathematics

1 answer:

Norma-Jean [14]4 years ago

3 0

Answer:

<em>x</em> = -1

Step-by-step explanation:

Let's solve your equation step-by-step.

7x + 12 = 2(1 − x) − x

Step 1: Simplify both sides of the equation.

7x + 12 = 2(1 − x) − x

7x + 12 = (2) (1) + (2) (− x) + − x (Distribute)

7x + 12 = 2 + − 2x + − x

7x + 12 = (− 2x + − x) + (2) (Combine Like Terms)

7x + 12 = − 3x + 2

Step 2: Add 3x to both sides.

7x + 12 + 3x = − 3x + 2 + 3x

10x + 12 = 2

Step 3: Subtract 12 from both sides.

10x + 12 − 12 = 2 − 12

10x = − 10

Step 4: Divide both sides by 10.

10x/10 = - 10/10

x = - 1

Hope this helps! Sorry if it is confusing!

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A) The equilibrium prices P1 and P2 for two goods satisfy the equations:

Allisa [31]

The value of P1 and P2 using the inverse matrix is 5 and 6 respectively.

<h3 />

The equilibrium prices of the three independent commodities using the inverse matrix are P1 = 4, P2 = 7, P3 = 8 respectively.

<h3>What is a matrix?</h3>

A matrix can be defined as a collection of integers(numbers that are either positive or negative) that are organized in rows and columns to construct a rectangular array. The numbers in this matrix system are referred to as elements.

To determine the values of P1 and P2 for the system of equations given by using an inverse matrix, we have:

9P1 + P2 = 51

3P1 +4P2 = 39

Representing the above data in matrix form, we have:

$\left[\begin{array}{cc}9&1\\3&4\\ \end{array}\right] \left[\begin{array}{c} \mathbf{P_1} \\ \mathbf{P_2}\\ \end{array}\right] = \left[\begin{array}{c} \mathbf{51} \\ \mathbf{39} \\ \end{array}\right]$ which is in the form AX = B

In order for us to determine the values of P1 and P2, Let take the inverse of A⁻¹ on both sides of the AX= B, we have:

$\mathbf{{A^{-1} AX} = A^{-1} B}$

X = A⁻¹ B

Let's start by finding A⁻¹;

$\mathbf{A = \left[\begin{array}{cc}9&1\\3&4\end{array}\right] }$

$\mathbf{A^{-1} = \dfrac{1}{36-3}\left[\begin{array}{cc}4&-1\\3&9\end{array}\right] }$

$\mathbf{A^{-1} = \dfrac{1}{33}\left[\begin{array}{cc}4&-1\\3&9\end{array}\right] }$

Now, Let's Find A⁻¹B;

$\mathbf{A^{-1}B = \dfrac{1}{33}\left[\begin{array}{cc}4&-1\\3&9\end{array}\right] \left[\begin{array}{c}51\\39\\ \end{array}\right] }$

$\mathbf{\implies \dfrac{1}{33}\left[\begin{array}{cc}204&-39\\-153&+351\end{array}\right] }$

$\mathbf{\implies \left[\begin{array}{c}\dfrac{165}{33}\\ \\ \dfrac{198}{33}\end{array}\right] }$

$\mathbf{\implies \left[\begin{array}{c}5\\ \\ 6\end{array}\right] }$

$\left[\begin{array}{c}\mathbf{P_1}\\ \mathbf{P_2}\end{array}\right]= \left[\begin{array}{c}5\\ 6 \end{array}\right] }$

Therefore, we can conclude that the value of P1 and P2 using the inverse matrix is 5 and 6 respectively.

To determine the equilibrium prices of the three independent commodities using the inverse matrix, we have:

P₁ + 2P₂ + 3P₃ = 42

2P₁ + P₂ + 4P₃ = 47

P₁ + 3P₂ + 4P₃ = 57

The matrix in AX = B form is computed as:

$\implies\left[ \begin{array}{ccc}1&2&3\\2&1&4\\1&3&4\end{array}\right] \left[\begin{array}{c}P_1\\P_2\\P_3 \end{array}\right] = \left[\begin{array}{c}42\\47\\ 57\end{array}\right]$

$\mathbf{A^{-1} = \dfrac{1}{|A|} \ (adj \ A)}$

$\mathbf{A^{-1} = \dfrac{1}{1(4-12) -2(8-4) +3(6-1)} \left[\begin{array}{ccc}-8&4&5\\1&1&-1\\5&2&-3\end{array}\right] }^1$

$\mathbf{A^{-1}B = -1 \left[\begin{array}{ccc}-8&1&5\\-4&1&2\\5&2&-3\end{array}\right] }\left[\begin{array}{c}42\\47\\57\end{array}\right]$

$\mathbf{A^{-1}B = -\left[\begin{array}{ccc}-336&+47&+285\\-168&+47&+114\\210&-47&-171\end{array}\right] }$

$\mathbf{A^{-1}B = -\left[\begin{array}{c}-4\\-7\\-8\end{array}\right] }$

$\mathbf{A^{-1}B = \left[\begin{array}{c}4\\7\\8\end{array}\right] }$

Therefore, we can conclude that the values of P1 = 4, P2 = 7, P3 = 8 respectively.

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Since it is between 7 and 9, it is 7+9=16. 16/2=8

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7x+12=2(1-x)-x solve​

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