SAT

## Linear equations in one variable A **linear equation** in one variable can be rearranged to $ax + b = c$, and solving it means undoing what has been done to $x$ in the reverse order: clear brackets and fractions first, collect the $x$ terms on one side, then divide. Solve $3(2x - 5) + 4 = 5x + 1$. $$ \begin{aligned} 6x - 15 + 4 &= 5x + 1 \\ 6x - 11 &= 5x + 1 \\ x &= 12 \end{aligned} $$ Checking in the original: the left side is $3(24 - 5) + 4 = 61$ and the right side is $60 + 1 = 61$. When an equation has fractions, multiplying every term by a common denominator removes them. For $\frac{x}{4} + \frac{x}{6} = 5$ the common denominator is 12, which gives $3x + 2x = 60$ and so $x = 12$. ### How many solutions Take an equation of the form $ax + b = cx + d$ and collect terms to get $(a - c)x = d - b$. There are three cases. | Condition | Result | What it looks like | |---|---|---| | $a \ne c$ | exactly one solution | $x = \frac{d - b}{a - c}$ | | $a = c$ and $b \ne d$ | no solution | a false statement such as $3 = 7$ | | $a = c$ and $b = d$ | infinitely many solutions | a true statement such as $6 = 6$ | This is how questions with an unknown constant are solved. If $4(x + 3) = kx + 12$ has infinitely many solutions, the left side is $4x + 12$, so the coefficients of $x$ must match and $k = 4$. If $6x + 5 = kx + 9$ has no solution, the $x$ coefficients must match while the constants differ, so $k = 6$. ### Rearranging a formula Solving for one letter in terms of the others…