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lana [24]
2 years ago
9

Solve 2(9-x) = 3 ( x +16 )

Mathematics
1 answer:
andreyandreev [35.5K]2 years ago
3 0

Answer: x=-6

Step-by-step explanation:

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

2(9−x)=3(x+16)

Step 1: Simplify both sides of the equation by distribution

2(9-x)=3(x+16)\\(2)(9)+(2)(-x)=(3)(x)+(3)(16)\\18+-2x=3x+48\\-2x+18=3x+48\\

Step 2: Subtract 3x from both sides.

-2x+18-3x=3x+48-3x\\-5x+18=48

Step 3: Subtract 18 from both sides.

-5x+18-18=48-18\\-5x=30

Step 4: Divide both sides by -5.

\frac{-5x}{-5} =\frac{30}{-5} \\x=-6

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Answer:

See attached photo for graph

Step-by-step explanation:

Solve the equation for y...  (that means get y by itself)

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y = -(5/4)x - 6/4  (divide both sides by 4)

y = -(5/4)x - 3/2      (simplify)

Now you can graph using slope-intercept method.   The y-intercept is -3/2, and the slope is -5/4    

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Given vectors u = (−1, 2, 3) and v = (3, 4, 2) in R 3 , consider the linear span: Span{u, v} := {αu + βv: α, β ∈ R}. Are the vec
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Answer:

(2,6,6) \not \in \text{Span}(u,v)

(-9,-2,5)\in \text{Span}(u,v)

Step-by-step explanation:

Let b=(b_1,b_2,b_3) \in \mathbb{R}^3. We have that b\in \text{Span}\{u,v\} if and only if we can find scalars \alpha,\beta \in \mathbb{R} such that \alpha u + \beta v = b. This can be translated to the following equations:

1. -\alpha + 3 \beta = b_1

2.2\alpha+4 \beta = b_2

3. 3 \alpha +2 \beta = b_3

Which is a system of 3 equations a 2 variables. We can take two of this equations, find the solutions for \alpha,\beta and check if the third equationd is fulfilled.

Case (2,6,6)

Using equations 1 and 2 we get

-\alpha + 3 \beta = 2

2\alpha+4 \beta = 6

whose unique solutions are \alpha =1 = \beta, but note that for this values, the third equation doesn't hold (3+2 = 5 \neq 6). So this vector is not in the generated space of u and v.

Case (-9,-2,5)

Using equations 1 and 2 we get

-\alpha + 3 \beta = -9

2\alpha+4 \beta = -2

whose unique solutions are \alpha=3, \beta=-2. Note that in this case, the third equation holds, since 3(3)+2(-2)=5. So this vector is in the generated space of u and v.

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2 years ago
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