Question

solve for x.
view solution history
add to both sides ;add what to both sides?
2x +
= 20 + 3x +

\frac{2}{5}x = 4 + \frac{3}{5}x
2x = 4 + 5 + 3x

Explanation:

Step1: Identify target terms to isolate $x$

We need to eliminate $3x$ from the right side and constants from the left. To cancel $3x$ on the right, add $-3x$ (or subtract $3x$) to both sides. To cancel the constant placeholder on the left, we can work backward from the sample equation $\frac{2}{5}x = 4 + \frac{3}{5}x$ (scaled version) which translates to $2x = 20 + 3x$ when multiplied by 5.

Step2: Apply inverse operation to both sides

To solve $2x + \square = 20 + 3x + \square$, we add $-3x$ to both sides to group $x$-terms, and add $-20$ to both sides to group constants (combined, we add $-3x -20$ to both sides, but for the blank structure:
Left blank: $-3x$ (to cancel $3x$ on the right)
Right blank: $-20$ (to cancel 20 on the left)
Or, to simplify the equation to standard form:
Add $-3x$ to both sides:
$2x + (-3x) = 20 + 3x + (-3x)$
$\implies -x = 20$

Step3: Solve for $x$

Multiply both sides by $-1$:
$x = -20$

Answer:

To balance the equation and solve for $x$, add $\boldsymbol{-3x}$ to the left blank and $\boldsymbol{-20}$ to the right blank (or add $\boldsymbol{-3x -20}$ to both sides). The final value of $x$ is $\boldsymbol{-20}$.