All categories

2 years ago

4x-5=x-1+3x I need help

Mathematics

1 answer:

labwork [276]2 years ago

8 0

Answer:

x = no solutions

General Formulas and Concepts:

Pre-Algebra

Order of Operations: BPEMDAS

Brackets
Parenthesis
Exponents
Multiplication
Division
Addition
Subtraction

Left to Right

Equality Properties

Step-by-step explanation:

Step 1: Define equation

4x - 5 = x - 1 + 3x

Step 2: Solve for x

Combine like terms: 4x - 5 = 4x - 1
Subtract 4x on both sides: -5 ≠ -1

Here we see that -5 does not equal -1.

∴ This equation has no solutions.

You might be interested in

Write an equation in slope-intercept form for the line that passes through the

Pavlova-9 [17]

Make slope intercept
3x - y = 5
3x = y + 5
y = 3x - 5
Parallel = same slope
Slope = 3
Y = 3x + b
Plug in the point
-2 = 3(-1) + b
-2 = -3 + b, b = 1

Solution: y = 3x + 1

6 0

2 years ago

Read 2 more answers

Describe its graph. 4 – 5x > 19

ira [324]

Answer:

X is greater than 4.6

Step-by-step explanation:

7 0

3 years ago

Determine if (-2,6) and (4,12) are solutions to the system of equations:

kykrilka [37]

Answer:

Both $(-2,6)$ and $(4,12)$ are solutions to the system.

Step-by-step explanation:

In order to determine whether the two given points represent solutions to our system of equations, we must "plug" thos points into both equations and check that the equality remains valid.

Step 1: Plug $(-2,6)$ into $y=x+8$

$(6) = (-2)+8 = 6$

The solution verifies the equation.

Step 2: Plug $(-2,6)$ into $2y=x^2 + 8$

$2(6) = (-2)^2+8=4+8=12$

The solution verifies both equations. Therefore, $(-2,6)$ is a solution to this system.

Now we must check if the second point is also valid.

Step 3: Plug $(4,12)$ into $y=x+8$

$(12) = (4)+8 = 12$

Step 4: Plug $(4,12)$ into $2y=x^2 + 8$

$2(12) = (4)^2+8=16+8=24$

The solution verifies both equations. Therefore, $(4,12)$ is another solution to this system.

3 0

2 years ago

Which system of linear equations could be used to determine the price of each book

GuDViN [60]

Answer:

Let the price of the maths book be m and price of the novel book be n

Given that,

Total cost of the books is $54

The price of math book is $8 more than 3 times the price of novel book.

we get,

The system of equation as,

$\begin{gathered} m+n=54 \\ m=8+3n \end{gathered}$

Hence the system of equation to determine the price of the maths and novel book is,

$\begin{gathered} m+n=54 \\ m=8+3n \end{gathered}$

4 0

1 year ago

Prove that if n is a perfect square then n + 2 is not a perfect square

notka56 [123]

Answer:

This statement can be proven by contradiction for $n \in \mathbb{N}$ (including the case where $n = 0$ .)

$\text{Let $n \in \mathbb{N}$ be a perfect square}$ .

$\textbf{Case 1.} ~ \text{n = 0}$ :

$\text{$n + 2 = 2$, which isn't a perfect square}$ .

$\text{Claim verified for $n = 0$}$ .

$\textbf{Case 2.} ~ \text{$n \in \mathbb{N}$ and $n \ne 0$. Hence $n \ge 1$}$ .

$\text{Assume that $n$ is a perfect square}$ .

$\text{$\iff$ $\exists$ $a \in \mathbb{N}$ s.t. $a^2 = n$}$ .

$\text{Assume $\textit{by contradiction}$ that $(n + 2)$ is a perfect square}$ .

$\text{$\iff$ $\exists$ $b \in \mathbb{N}$ s.t. $b^2 = n + 2$}$ .

$\text{$n + 2 > n > 0$ $\implies$ $b = \sqrt{n + 2} > \sqrt{n} = a$}$ .

$\text{$a,\, b \in \mathbb{N} \subset \mathbb{Z}$ $\implies b - a = b + (- a) \in \mathbb{Z}$}$ .

$\text{$b > a \implies b - a > 0$. Therefore, $b - a \ge 1$}$ .

$\text{$\implies b \ge a + 1$}$ .

$\text{$\implies n+ 2 = b^2 \ge (a + 1)^2= a^2 + 2\, a + 1 = n + 2\, a + 1$}$ .

$\text{$\iff 1 \ge 2\,a $}$ .

$\text{$\displaystyle \iff a \le \frac{1}{2}$}$ .

$\text{Contradiction (with the assumption that $a \ge 1$)}$ .

$\text{Hence the original claim is verified for $n \in \mathbb{N}\backslash\{0\}$}$ .

$\text{Hence the claim is true for all $n \in \mathbb{N}$}$ .

Step-by-step explanation:

Assume that the natural number $n \in \mathbb{N}$ is a perfect square. Then, (by the definition of perfect squares) there should exist a natural number $a$ ( $a \in \mathbb{N}$ ) such that $a^2 = n$ .

Assume by contradiction that $n + 2$ is indeed a perfect square. Then there should exist another natural number $b \in \mathbb{N}$ such that $b^2 = (n + 2)$ .

Note, that since $(n + 2) > n \ge 0$ , $\sqrt{n + 2} > \sqrt{n}$ . Since $b = \sqrt{n + 2}$ while $a = \sqrt{n}$ , one can conclude that $b > a$ .

Keep in mind that both $a$ and $b$ are natural numbers. The minimum separation between two natural numbers is $1$ . In other words, if $b > a$ , then it must be true that $b \ge a + 1$ .

Take the square of both sides, and the inequality should still be true. (To do so, start by multiplying both sides by $(a + 1)$ and use the fact that $b \ge a + 1$ to make the left-hand side $b^2$ .)

$b^2 \ge (a + 1)^2$ .

Expand the right-hand side using the binomial theorem:

$(a + 1)^2 = a^2 + 2\,a + 1$ .

$b^2 \ge a^2 + 2\,a + 1$ .

However, recall that it was assumed that $a^2 = n$ and $b^2 = n + 2$ . Therefore,

$\underbrace{b^2}_{=n + 2)} \ge \underbrace{a^2}_{=n} + 2\,a + 1$ .

$n + 2 \ge n + 2\, a + 1$ .

Subtract $n + 1$ from both sides of the inequality:

$1 \ge 2\, a$ .

$\displaystyle a \le \frac{1}{2} = 0.5$ .

Recall that $a$ was assumed to be a natural number. In other words, $a \ge 0$ and $a$ must be an integer. Hence, the only possible value of $a$ would be $0$ .

Since $a$ could be equal $0$ , there's not yet a valid contradiction. To produce the contradiction and complete the proof, it would be necessary to show that $a = 0$ just won't work as in the assumption.

If indeed $a = 0$ , then $n = a^2 = 0$ . $n + 2 = 2$ , which isn't a perfect square. That contradicts the assumption that if $n = 0$ is a perfect square, $n + 2 = 2$ would be a perfect square. Hence, by contradiction, one can conclude that

$\text{if $n$ is a perfect square, then $n + 2$ is not a perfect square.}$ .

Note that to produce a more well-rounded proof, it would likely be helpful to go back to the beginning of the proof, and show that $n \ne 0$ . Then one can assume without loss of generality that $n \ne 0$ . In that case, the fact that $\displaystyle a \le \frac{1}{2}$ is good enough to count as a contradiction.

7 0

3 years ago