Set Up Equations: We have a system of two linear equations: 1) 2x - 3y = -1 2) y = x - 1 Let's solve the second equation for y and substitute it into the first equation.Substitute and Simplify: Substitute y from the second equation into the first equation: 2x - 3(x - 1) = -1 Now, let's simplify and solve for x.Isolate x: Distribute the -3 across the parentheses: 2x - 3x + 3 = -1 Combine like terms: -x + 3 = -1Find y: Add x to both sides to isolate the constant on the right side: 3 = x - 1 Now, solve for x by adding 1 to both sides: x = 3 + 1 x = 4Final Solution: Now that we have the value of x, we can substitute it back into the second equation to find y: y = x - 1 y = 4 - 1 y = 3

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$2x-3y=-1$ $\newline$ $y=x-1$

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Precalculus

Solve logarithmic equations with multiple logarithms

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2x-3y=-1

\newline

y=x-1

Set Up Equations: We have a system of two linear equations: $\newline$ $1$ ) $2x - 3y = -1$ $\newline$ $2$ ) $y = x - 1$ $\newline$ Let's solve the second equation for $y$ and substitute it into the first equation.
Substitute and Simplify: Substitute $y$ from the second equation into the first equation: $2x - 3(x - 1) = -1$ Now, let's simplify and solve for $x$ .
Isolate $x$ : Distribute the $-3$ across the parentheses: $2x - 3x + 3 = -1$ Combine like terms: $-x + 3 = -1$
Find $y$ : Add $x$ to both sides to isolate the constant on the right side: $\newline$ $3 = x - 1$ $\newline$ Now, solve for $x$ by adding $1$ to both sides: $\newline$ $x = 3 + 1$ $\newline$ $x = 4$
Final Solution: Now that we have the value of $x$ , we can substitute it back into the second equation to find $y$ : $y = x - 1$ $y = 4 - 1$ $y = 3$