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3 years ago

Whats the answer for 6b+7-2b=1+5b

Mathematics

2 answers:

Alecsey [184]3 years ago

3 0

$Solution,6b+7-2b=1+5b\quad :\quad b=6$

$Steps:$

$6b+7-2b=1+5b$

$\mathrm{Group\:like\:terms}, 6b-2b+7=1+5b$

$\mathrm{Add\:similar\:elements:}\:6b-2b=4b, 4b+7=1+5b$

$\mathrm{Subtract\:}7\mathrm{\:from\:both\:sides}, 4b+7-7=1+5b-7$

$\mathrm{Simplify}, 4b=5b-6$

$\mathrm{Subtract\:}5b\mathrm{\:from\:both\:sides}, 4b-5b=5b-6-5b$

$\mathrm{Simplify}, -b=-6$

$\mathrm{Divide\:both\:sides\:by\:}-1, \frac{-b}{-1}=\frac{-6}{-1}$

$\mathrm{Simplify}, b=6$

$\mathrm{The\:Correct\:Answer\:is\:b=6}$

$\mathrm{Hope\:This\:Helps!!!}$

$\mathrm{-Austint1414}$

MrMuchimi3 years ago

3 0

Answer:

b=6

Step-by-step explanation:

We have the expression $6b+7-2b=1+5b$

We are to leave in one side of the expression all the terms that has <em>b</em> and in the other the terms that doesn't have <em>b.</em>

$6b+7-2b=1+5b\\6b-2b-5b=1-7$

We can apply common factor <em>b </em>in the left side, and resolve the subtraction in the right side.

$b(6-2-5)=-6\\b(-1)=-6$

Now divide both sides in (-1).

$b(-1)=-6\\\frac{b(-1)}{(-1)}=\frac{(-6)}{(-1)} \\b=6$

Then the answer of $6b+7-2b=1+5b$ is b=6.

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The larger of two numbers is five more than twice the smaller number. The sum of the two numbers is 38. Find the numbers.

Damm [24]

Answer:

I would set this up as:

let: lesser number = x

greater number = 2x + 5

Restating the word problem:

lesser number + greater number = 38

x + 2x + 5 = 38

Solving: 3x + 5 = 38

- 5 -5

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

3x = 33

--- ---

3 3

x = 11

Substitution for the greater number:

2(11) + 5

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8 0

3 years ago

Read 2 more answers

What is the equation for the plane illustrated below?

TiliK225 [7]

Answer:

Hence, none of the options presented are valid. The plane is represented by $3 \cdot x + 3\cdot y + 2\cdot z = 6$ .

Step-by-step explanation:

The general equation in rectangular form for a 3-dimension plane is represented by:

$a\cdot x + b\cdot y + c\cdot z = d$

Where:

$x$ , $y$ , $z$ - Orthogonal inputs.

$a$ , $b$ , $c$ , $d$ - Plane constants.

The plane presented in the figure contains the following three points: (2, 0, 0), (0, 2, 0), (0, 0, 3)

For the determination of the resultant equation, three equations of line in three distinct planes orthogonal to each other. That is, expressions for the xy, yz and xz-planes with the resource of the general equation of the line:

xy-plane (2, 0, 0) and (0, 2, 0)

$y = m\cdot x + b$

$m = \frac{y_{2}-y_{1}}{x_{2}-x_{1}}$

Where:

$m$ - Slope, dimensionless.

$x_{1}$ , $x_{2}$ - Initial and final values for the independent variable, dimensionless.

$y_{1}$ , $y_{2}$ - Initial and final values for the dependent variable, dimensionless.

$b$ - x-Intercept, dimensionless.

If $x_{1} = 2$ , $y_{1} = 0$ , $x_{2} = 0$ and $y_{2} = 2$ , then:

Slope

$m = \frac{2-0}{0-2}$

$m = -1$

x-Intercept

$b = y_{1} - m\cdot x_{1}$

$b = 0 -(-1)\cdot (2)$

$b = 2$

The equation of the line in the xy-plane is $y = -x+2$ or $x + y = 2$ , which is equivalent to $3\cdot x + 3\cdot y = 6$ .

yz-plane (0, 2, 0) and (0, 0, 3)

$z = m\cdot y + b$

$m = \frac{z_{2}-z_{1}}{y_{2}-y_{1}}$

Where:

$m$ - Slope, dimensionless.

$y_{1}$ , $y_{2}$ - Initial and final values for the independent variable, dimensionless.

$z_{1}$ , $z_{2}$ - Initial and final values for the dependent variable, dimensionless.

$b$ - y-Intercept, dimensionless.

If $y_{1} = 2$ , $z_{1} = 0$ , $y_{2} = 0$ and $z_{2} = 3$ , then:

Slope

$m = \frac{3-0}{0-2}$

$m = -\frac{3}{2}$

y-Intercept

$b = z_{1} - m\cdot y_{1}$

$b = 0 -\left(-\frac{3}{2} \right)\cdot (2)$

$b = 3$

The equation of the line in the yz-plane is $z = -\frac{3}{2}\cdot y+3$ or $3\cdot y + 2\cdot z = 6$ .

xz-plane (2, 0, 0) and (0, 0, 3)

$z = m\cdot x + b$

$m = \frac{z_{2}-z_{1}}{x_{2}-x_{1}}$

Where:

$m$ - Slope, dimensionless.

$x_{1}$ , $x_{2}$ - Initial and final values for the independent variable, dimensionless.

$z_{1}$ , $z_{2}$ - Initial and final values for the dependent variable, dimensionless.

$b$ - z-Intercept, dimensionless.

If $x_{1} = 2$ , $z_{1} = 0$ , $x_{2} = 0$ and $z_{2} = 3$ , then:

Slope

$m = \frac{3-0}{0-2}$

$m = -\frac{3}{2}$

x-Intercept

$b = z_{1} - m\cdot x_{1}$

$b = 0 -\left(-\frac{3}{2} \right)\cdot (2)$

$b = 3$

The equation of the line in the xz-plane is $z = -\frac{3}{2}\cdot x+3$ or $3\cdot x + 2\cdot z = 6$

After comparing each equation of the line to the definition of the equation of the plane, the following coefficients are obtained:

$a = 3$ , $b = 3$ , $c = 2$ , $d = 6$

Hence, none of the options presented are valid. The plane is represented by $3 \cdot x + 3\cdot y + 2\cdot z = 6$ .

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3 years ago

What is -x divided by 4?

exis [7]

The answer is just -x/4 because you can divide a variable

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3 years ago

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Which value, when placed in the box, would result in a system of equations with infinitely many solutions? y = -2x 4 6x 3y =

wariber [46]

The missing value is 12 in a system of equations with infinitely many solutions conditions.

It is given that in the system of equations there are two equations given:

$\rm y =n -2x+4\\\rm and \\\rm 6x+3y= ?$

It is required to find the missing value in the second equation.

<h3>What is a linear 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.

We have equations:

$\rm y = -2x+4 ....(1)\\ \\\rm 6x+3y= ? ....(2)$

Let's suppose the missing value is 'Z'

We know that the two pairs of equations have infinitely many solutions if and if they have the same coefficients of variables and the same constant on both sides.

From equation (1)

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