$1=\sqrt{3 x+1}-\sqrt{2 x-1}$

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Algebra 1

Domain and range of absolute value functions: equations

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1=\sqrt{3 x+1}-\sqrt{2 x-1}

Given Equation: We start with the given equation: $\newline$ $1 = \sqrt{3x+1} - \sqrt{2x-1}$ $\newline$ To solve for $x$ , we need to isolate the square root terms and then square both sides to eliminate the square roots.
Move and Square: First, we move the $\sqrt{2x-1}$ to the other side of the equation by adding it to both sides: $1 + \sqrt{2x-1} = \sqrt{3x+1}$
Expand and Simplify: Now, we square both sides of the equation to eliminate the square roots: $\newline$ (\(1 + \sqrt{ $2$ x $-1$ })^ $2$ = (\sqrt{ $3$ x+ $1$ })^ $2$
Combine Like Terms: Expanding the left side, we get: $1 + 2\sqrt{2x-1} + (\sqrt{2x-1})^2 = 3x+1$
Isolate Square Root: Simplify the left side by squaring $\sqrt{2x-1}$ : $1 + 2\sqrt{2x-1} + 2x-1 = 3x+1$
Square Both Sides: Combine like terms on the left side: $2x + 2\sqrt{2x-1} = 3x+1$
Clear Fraction: Subtract $2x$ from both sides to get the square root term by itself: $\newline$ $2\sqrt{2x-1} = x+1$
Expand and Rearrange: Divide both sides by $2$ to isolate the square root: $\sqrt{2x-1} = \frac{x+1}{2}$
Factor Quadratic: Now, square both sides again to eliminate the square root: \sqrt{$2$x$-1$})^\(2 = \left(\frac{x+ $1$ }{ $2$ }\right)^ $2$
Check Solutions: Squaring the left side gives us $2x-1$ , and expanding the right side gives us: $\newline$ $2x-1 = \frac{x^2 + 2x + 1}{4}$
Check Solutions: Squaring the left side gives us $2x-1$ , and expanding the right side gives us: $\newline$ $2x-1 = \frac{x^2 + 2x + 1}{4}$ Multiply both sides by $4$ to clear the fraction: $\newline$ $4(2x-1) = x^2 + 2x + 1$
Check Solutions: Squaring the left side gives us $2x-1$ , and expanding the right side gives us: $\newline$ $2x-1 = \frac{x^2 + 2x + 1}{4}$ Multiply both sides by $4$ to clear the fraction: $\newline$ $4(2x-1) = x^2 + 2x + 1$ Expand the left side: $\newline$ $8x - 4 = x^2 + 2x + 1$
Check Solutions: Squaring the left side gives us $2x-1$ , and expanding the right side gives us: $\newline$ $2x-1 = \frac{x^2 + 2x + 1}{4}$ Multiply both sides by $4$ to clear the fraction: $\newline$ $4(2x-1) = x^2 + 2x + 1$ Expand the left side: $\newline$ $8x - 4 = x^2 + 2x + 1$ Rearrange the equation to set it to zero and form a quadratic equation: $\newline$ $x^2 - 6x + 5 = 0$
Check Solutions: Squaring the left side gives us $2x-1$ , and expanding the right side gives us: $\newline$ $2x-1 = \frac{x^2 + 2x + 1}{4}$ Multiply both sides by $4$ to clear the fraction: $\newline$ $4(2x-1) = x^2 + 2x + 1$ Expand the left side: $\newline$ $8x - 4 = x^2 + 2x + 1$ Rearrange the equation to set it to zero and form a quadratic equation: $\newline$ $x^2 - 6x + 5 = 0$ Factor the quadratic equation: $\newline$ $(x - 5)(x - 1) = 0$
Check Solutions: Squaring the left side gives us $2x-1$ , and expanding the right side gives us: $\newline$ $2x-1 = \frac{x^2 + 2x + 1}{4}$ Multiply both sides by $4$ to clear the fraction: $\newline$ $4(2x-1) = x^2 + 2x + 1$ Expand the left side: $\newline$ $8x - 4 = x^2 + 2x + 1$ Rearrange the equation to set it to zero and form a quadratic equation: $\newline$ $x^2 - 6x + 5 = 0$ Factor the quadratic equation: $\newline$ $(x - 5)(x - 1) = 0$ Set each factor equal to zero and solve for $x$ : $\newline$ $x - 5 = 0$ or $x - 1 = 0$
Check Solutions: Squaring the left side gives us $2x-1$ , and expanding the right side gives us: $\newline$ $2x-1 = \frac{x^2 + 2x + 1}{4}$ Multiply both sides by $4$ to clear the fraction: $\newline$ $4(2x-1) = x^2 + 2x + 1$ Expand the left side: $\newline$ $8x - 4 = x^2 + 2x + 1$ Rearrange the equation to set it to zero and form a quadratic equation: $\newline$ $x^2 - 6x + 5 = 0$ Factor the quadratic equation: $\newline$ $(x - 5)(x - 1) = 0$ Set each factor equal to zero and solve for $x$ : $\newline$ $x - 5 = 0$ or $x - 1 = 0$ Solving each equation gives us the possible solutions for $x$ : $\newline$ $2x-1 = \frac{x^2 + 2x + 1}{4}$ $1$ or $2x-1 = \frac{x^2 + 2x + 1}{4}$ $2$
Check Solutions: Squaring the left side gives us $2x-1$ , and expanding the right side gives us: $\newline$ $2x-1 = \frac{x^2 + 2x + 1}{4}$ Multiply both sides by $4$ to clear the fraction: $\newline$ $4(2x-1) = x^2 + 2x + 1$ Expand the left side: $\newline$ $8x - 4 = x^2 + 2x + 1$ Rearrange the equation to set it to zero and form a quadratic equation: $\newline$ $x^2 - 6x + 5 = 0$ Factor the quadratic equation: $\newline$ $(x - 5)(x - 1) = 0$ Set each factor equal to zero and solve for $x$ : $\newline$ $x - 5 = 0$ or $x - 1 = 0$ Solving each equation gives us the possible solutions for $x$ : $\newline$ $2x-1 = \frac{x^2 + 2x + 1}{4}$ $1$ or $2x-1 = \frac{x^2 + 2x + 1}{4}$ $2$ We need to check these solutions in the original equation because squaring both sides can introduce extraneous solutions. Let's check $2x-1 = \frac{x^2 + 2x + 1}{4}$ $1$ : $\newline$ $2x-1 = \frac{x^2 + 2x + 1}{4}$ $4$ $\newline$ $2x-1 = \frac{x^2 + 2x + 1}{4}$ $5$ $\newline$ $2x-1 = \frac{x^2 + 2x + 1}{4}$ $6$ $\newline$ $2x-1 = \frac{x^2 + 2x + 1}{4}$ $7$ $\newline$ This solution checks out.
Check Solutions: Squaring the left side gives us $2x-1$ , and expanding the right side gives us: $\newline$ $2x-1 = \frac{x^2 + 2x + 1}{4}$ Multiply both sides by $4$ to clear the fraction: $\newline$ $4(2x-1) = x^2 + 2x + 1$ Expand the left side: $\newline$ $8x - 4 = x^2 + 2x + 1$ Rearrange the equation to set it to zero and form a quadratic equation: $\newline$ $x^2 - 6x + 5 = 0$ Factor the quadratic equation: $\newline$ $(x - 5)(x - 1) = 0$ Set each factor equal to zero and solve for $x$ : $\newline$ $x - 5 = 0$ or $x - 1 = 0$ Solving each equation gives us the possible solutions for $x$ : $\newline$ $2x-1 = \frac{x^2 + 2x + 1}{4}$ $1$ or $2x-1 = \frac{x^2 + 2x + 1}{4}$ $2$ We need to check these solutions in the original equation because squaring both sides can introduce extraneous solutions. Let's check $2x-1 = \frac{x^2 + 2x + 1}{4}$ $1$ : $\newline$ $2x-1 = \frac{x^2 + 2x + 1}{4}$ $4$ $\newline$ $2x-1 = \frac{x^2 + 2x + 1}{4}$ $5$ $\newline$ $2x-1 = \frac{x^2 + 2x + 1}{4}$ $6$ $\newline$ $2x-1 = \frac{x^2 + 2x + 1}{4}$ $7$ $\newline$ This solution checks out.Now let's check $2x-1 = \frac{x^2 + 2x + 1}{4}$ $2$ : $\newline$ $2x-1 = \frac{x^2 + 2x + 1}{4}$ $9$ $\newline$ $4$ $0$ $\newline$ $4$ $1$ $\newline$ $2x-1 = \frac{x^2 + 2x + 1}{4}$ $7$ $\newline$ This solution also checks out.