Solve 1 - (x - 2) - [(x - 3) - (x - 1)] = 0 | Equatix - Math Solver

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Example

3(r+2s)=2t-4

a⋅(b-2)=3b

3(r+2s)=2t-4

a⋅(b-2)=3b

Question

$$1-(x-2)-[(x-3)-(x-1)]=0.$$

Solve for x

$x=5$

Steps for Solving Linear Equation

To find the opposite of $x-2$, find the opposite of each term.

$$1-x-\left(-2\right)-\left(x-3-\left(x-1\right)\right)=0$$

The opposite of $-2$ is $2$.

$$1-x+2-\left(x-3-\left(x-1\right)\right)=0$$

Add $1$ and $2$ to get $3$.

$$3-x-\left(x-3-\left(x-1\right)\right)=0$$

To find the opposite of $x-1$, find the opposite of each term.

$$3-x-\left(x-3-x-\left(-1\right)\right)=0$$

The opposite of $-1$ is $1$.

$$3-x-\left(x-3-x+1\right)=0$$

Combine $x$ and $-x$ to get $0$.

$$3-x-\left(-3+1\right)=0$$

Add $-3$ and $1$ to get $-2$.

$$3-x-\left(-2\right)=0$$

The opposite of $-2$ is $2$.

$$3-x+2=0$$

Add $3$ and $2$ to get $5$.

$$5-x=0$$

Subtract $5$ from both sides. Anything subtracted from zero gives its negation.

$$-x=-5$$

Multiply both sides by $-1$.

$$x=5$$