Solving Equations by using Addition/ Subtraction: Test if the coefficient is different from 1

To solve for $x$, first, we need to get all terms involving $x$ on one side of the equation and constant terms on the other. Start with the original equation:

$2x + 4 = 3x - 5$

Subtract $2x$ from both sides to isolate the term involving $x$ on one side:

$4 = x - 5$

Next, add 5 to both sides to isolate $x$:

$9 = x$

Thus, the value of $x$ is $9$.

To solve the equation $3x + 5 = 2x + 20$, we need to find the value of $x$ that satisfies this equation. Here are the detailed steps:

• Step 1: Eliminate the variable from one side.
  We want to get all terms involving $x$ on one side and constant terms on the other side. First, subtract $2x$ from both sides of the equation to eliminate $x$ from the right side.

  $3x + 5 - 2x = 2x + 20 - 2x$

  This simplifies to:

  $x + 5 = 20$

• Step 2: Simplify the equation.
  Now, we need to isolate $x$ by removing the constant term from the left side. Subtract 5 from both sides:

  $x + 5 - 5 = 20 - 5$

  This simplifies to:

  $x = 15$

• Step 3: Verify the solution.
  Substitute $x = 15$ back into the original equation to check if it holds true:

  $3(15) + 5 = 2(15) + 20$

  This results in:

  $45 + 5 = 30 + 20$

  $50 = 50$

  Since both sides of the equation are equal,$x = 15$ is indeed the correct solution.

Therefore, the solution to the equation $3x + 5 = 2x + 20$ is $x = 15$.

We start with the equation:
$4x + 4 = 5x + 2$

Our goal is to solve for $x$. To do this, we aim to collect all terms containing $x$ on one side of the equation and constant terms on the other side. First, subtract $4x$ from both sides of the equation to eliminate the $x$ term on the left side:

$4x + 4 - 4x = 5x + 2 - 4x$

This simplifies the equation to:

$4 = x + 2$

Next, subtract $2$ from both sides to isolate the variable $x$ on the right side:

$4 - 2 = x + 2 - 2$

This gives us:

$2 = x$

Thus, the solution to the equation is $x = 2$.

To solve the equation $5x + 2 = 4x + 10$, we can simplify and solve for $x$ by following these steps:

• First, let's get all terms involving $x$ on one side and the constant terms on the other. We do this by subtracting $4x$ from both sides:

  $5x + 2 - 4x = 4x + 10 - 4x$

  This simplifies to:

  $x + 2 = 10$

• Next, we need to isolate $x$ by subtracting 2 from both sides:

  $x + 2 - 2 = 10 - 2$

  Which simplifies to:

  $x = 8$

Thus, the solution for $x$ is $8$.

The given equation is: $6x-3=7x+5$

Our goal is to solve for $x$. To achieve this, we'll first get all the terms containing $x$ on one side of the equation and constants on the other side.

Step 1: Subtract $6x$ from both sides to get all $x$ terms on one side:

• $6x - 3 - 6x = 7x + 5 - 6x$

This simplifies to:

• $-3 = x + 5$

Step 2: Next, subtract $5$ from both sides to isolate $x$:

• $-3 - 5 = x + 5 - 5$

This simplifies to:

• $-8 = x$

Therefore, the solution for $x$ is $-8$.

Start by moving the $7x$ term to the left side by subtracting $7x$ from both sides:
$8x - 7x - 1 = 7x + 5 - 7x$
This simplifies to:
$x - 1 = 5$

Next, add$1$ to both sides to isolate $x$:
$x - 1 + 1 = 5 + 1$
Simplifying this, we get:
$x = 6$.

To solve the equation $9x - 3 = 10x + 1$, we need to get all terms with $x$ on one side and constant terms on the other side. Here's how we do it step-by-step:

• First, subtract $9x$ from both sides of the equation to start getting $x$ terms on one side. This gives us: $-3 = x + 1$

• Next, subtract 1 from both sides to isolate $x$. We get: $-3 - 1 = x$

• Simplifying the left side, we find: $x = -4$

Therefore, the solution is $x = -4$.

To solve for$x$, first, get all terms involving $x$ on one side and constants on the other. Start from:

$4x - 7 = x + 5$

Subtract $x$ from both sides to simplify:

$3x - 7 = 5$

Add 7 to both sides to isolate the terms with$x$:

$3x = 12$

Divide each side by 3 to solve for$x$:

$x = 4$

Thus, $x$ is $4$.

First, we arrange the two sections so that the right side contains the values with the coefficient x and the left side the numbers without the x

Let's remember to maintain the plus and minus signs accordingly when we move terms between the sections.

First, we move a$5x$ to the right section and then the 22 to the left side. We obtain the following equation:

$-8-22=10x-5x$

We subtract both sides accordingly and obtain the following equation:

$-30=5x$

We divide both sections by 5 and obtain:

$-6=x$

We begin by placing parentheses around the two multiplication exercises:

$(20\times y)+(8\times2)-7=14$

We then solve the exercises within the parentheses:

$20y+16-7=14$

We simplify:

$20y+9=14$

We move the sections:

$20y=14-9$

$20y=5$

We divide by 20:

$y=\frac{5}{20}$

$y=\frac{5}{5\times4}$

We simplify:

$y=\frac{1}{4}$

To solve this exercise, we first need to identify that we have an equation with an unknown,

To solve such equations, the first step will be to arrange the equation so that on one side we have the numbers and on the other side the unknowns.

$2X+7-5X-12=-8X+3$

First, we'll move all unknowns to one side.
It's important to remember that when moving terms, the sign of the number changes (from negative to positive or vice versa).

$2X+7-5X-12+8X=3$

Now we'll do the same thing with the regular numbers.

$2X-5X+8X=3-7+12$

In the next step, we'll calculate the numbers according to the addition and subtraction signs.

$2X-5X=-3X$
$-3X+8X=5X$

$3-7=-4$$$
$-4+12=8$

$5X=8$

At this stage, we want to get to a state where we have only one $X$, not $5X$,
so we'll divide both sides of the equation by the coefficient of the unknown (in this case - 5).

$X={8\over5}$

It's important to remember that when we have regular numbers and unknowns, we cannot add or subtract them directly.

Let's collect like terms:

5b+2b-7+14=0

7b+7 = 0

Let's move terms

7b = -7

Let's divide by 7

b=-1

And that's the solution!