How do you write and solve the following? 5 less than the quotient of x and three is two

Staffangz
2022-09-12
Answered

How do you write and solve the following? 5 less than the quotient of x and three is two

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Christina Matthews

Answered 2022-09-13
Author has **16** answers

To solve:

Add 5 to both sides

$\frac{x}{3}-5+5=2+5$

$\frac{x}{3}=7$

Multiply both sides by 3

$\frac{x}{3}\cdot 3=7\cdot 3$

$x=21$

Add 5 to both sides

$\frac{x}{3}-5+5=2+5$

$\frac{x}{3}=7$

Multiply both sides by 3

$\frac{x}{3}\cdot 3=7\cdot 3$

$x=21$

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$\left|\begin{array}{ccc}1& -2& 5\\ 4& -5& 8\\ -3& 3& -3\end{array}\right|\left|\begin{array}{c}{x}_{1}\\ {x}_{2}\\ {x}_{3}\end{array}\right|=\left|\begin{array}{c}{b}_{1}\\ {b}_{2}\\ {b}_{3}\end{array}\right|$

I need to determine the values of the $b$ constants that would guarantee that the linear system is consistent. I tried to find the inverse of the matrix on the left hand side so that I could try and solve for the $x$ variables and see if there are any values of $b$ that would cause the system to be inconsistent but the matrix is singular. I then went on to put the matrix in reduced row echelon form.

$\left|\begin{array}{ccc}1& 0& -3\\ 0& 1& -4\\ 0& 0& 0\end{array}\right|$

From this I was able to derive the equations: ${x}_{1}-3{x}_{3}={b}_{1}$, ${x}_{2}-4{x}_{3}={b}_{2}$, and $0={b}_{3}$. I know that the answer is ${b}_{1}={b}_{2}+{b}_{3}$, but I don't know how to get that with the given information.

I need to determine the values of the $b$ constants that would guarantee that the linear system is consistent. I tried to find the inverse of the matrix on the left hand side so that I could try and solve for the $x$ variables and see if there are any values of $b$ that would cause the system to be inconsistent but the matrix is singular. I then went on to put the matrix in reduced row echelon form.

$\left|\begin{array}{ccc}1& 0& -3\\ 0& 1& -4\\ 0& 0& 0\end{array}\right|$

From this I was able to derive the equations: ${x}_{1}-3{x}_{3}={b}_{1}$, ${x}_{2}-4{x}_{3}={b}_{2}$, and $0={b}_{3}$. I know that the answer is ${b}_{1}={b}_{2}+{b}_{3}$, but I don't know how to get that with the given information.

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b.none of the given answers is true

c.$y}_{1}=t,{y}_{2}={e}^{-3t$

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