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 $x = 15$.

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:

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

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!