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3 years ago

Solve a=b+3/c-2 for c

Mathematics

1 answer:

Ronch [10]3 years ago

7 0

Answer:

c = 3 / a − b + 2

Explanation:

[ Step 1: Multiply both sides by c ]

ac = bc − 2c + 3

[ Step 2: Add -bc to both sides ]

ac + −bc = bc − 2c + 3 + −bc

ac − bc = −2c + 3

[ Step 3: Add 2c to both sides ]

ac − bc + 2c = −2c + 3 + 2c

ac − bc + 2c = 3

[ Step 4: Factor out variable c ]

c(a − b + 2) = 3

[ Step 5: Divide both sides by a - b + 2 ]

c(a − b + 2) / a − b + 2 = 3 / a − b + 2

c = 3 / a − b + 2

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Please help solve for x..

aalyn [17]

Answer:

x = 1 or x = -1

Step-by-step explanation:

Given equation:

$x^{100}-4^x \cdot x^{98}-x^2+4^x=0$

Factor out -1:

$\implies -1(-x^{100}+4^x \cdot x^{98}+x^2-4^x)=0$

Divide both sides by -1:

$\implies -x^{100}+4^x \cdot x^{98}+x^2-4^x=0$

Rearrange the terms:

$\implies 4^x \cdot x^{98}-4^x-x^{100}+x^2=0$

$\textsf{Apply exponent rule} \quad a^{b+c}=a^b \cdot a^c \quad \textsf{to }x^{100}:$

$\implies 4^x \cdot x^{98}-4^x-x^{98}x^2+x^2=0$

Factor the first two terms and the last two terms separately:

$\implies 4^x(x^{98}-1)-x^2(x^{98}-1)=0$

Factor out the common term $(x^{98}-1)$ :

$\implies (4^x-x^2)(x^{98}-1)=0$

<u>Zero Product Property</u>: If a ⋅ b = 0 then either a = 0 or b = 0 (or both).

Using the <u>Zero Product Property</u>, set each factor equal to zero and solve for x (if possible):

$\begin{aligned}x^{98}-1 & = 0 & \quad \textsf{or} \quad \quad4^x-x^2 & = 0 \\x^{98} & =1 & 4^x & = x^2 \\x & = 1, -1 & \textsf{no}& \textsf{ solutions for } x \in \mathbb{R}\end{aligned}$

Therefore, the solutions to the given equation are: x = 1 or x = -1

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