Solution 2.3:3d
From Förberedande kurs i matematik 1
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- | {{ | + | Because both terms, <math>x(x+3)</math> and <math>x(2x-9)</math>, contain the factor <math>x</math>, we can take out <math>x</math> from the left-hand side and collect together the remaining expression, |
- | < | + | |
- | {{ | + | {{Displayed math||<math>\begin{align} |
+ | x(x+3)-x(2x-9) | ||
+ | &= x\bigl((x+3)-(2x-9)\bigr)\\[5pt] | ||
+ | &= x(x+3-2x+9)\\[5pt] | ||
+ | &= x(-x+12)\,\textrm{.} | ||
+ | \end{align}</math>}} | ||
+ | |||
+ | The equation is thus | ||
+ | |||
+ | {{Displayed math||<math>x(-x+12) = 0</math>}} | ||
+ | |||
+ | and we obtain directly that the equation is satisfied if either <math>x</math> or <math>-x+12</math> is zero. The solutions to the equation are therefore <math>x=0</math> and <math>x=12</math>. | ||
+ | |||
+ | Here, it can be worth checking that <math>x=12</math> is a solution (the case | ||
+ | <math>x=0</math> is obvious) | ||
+ | |||
+ | {{Displayed math||<math>\text{LHS} = 12\cdot (12+3) - 12\cdot (2\cdot 12-9) = 2\cdot 15 - 12\cdot 15 = 0 = \text{RHS.}</math>}} |
Current revision
Because both terms, \displaystyle x(x+3) and \displaystyle x(2x-9), contain the factor \displaystyle x, we can take out \displaystyle x from the left-hand side and collect together the remaining expression,
\displaystyle \begin{align}
x(x+3)-x(2x-9) &= x\bigl((x+3)-(2x-9)\bigr)\\[5pt] &= x(x+3-2x+9)\\[5pt] &= x(-x+12)\,\textrm{.} \end{align} |
The equation is thus
\displaystyle x(-x+12) = 0 |
and we obtain directly that the equation is satisfied if either \displaystyle x or \displaystyle -x+12 is zero. The solutions to the equation are therefore \displaystyle x=0 and \displaystyle x=12.
Here, it can be worth checking that \displaystyle x=12 is a solution (the case \displaystyle x=0 is obvious)
\displaystyle \text{LHS} = 12\cdot (12+3) - 12\cdot (2\cdot 12-9) = 2\cdot 15 - 12\cdot 15 = 0 = \text{RHS.} |