解方程 $2(x-5)=3x-1$。 答案 $x$ = ... 解不等式 $7-4y>3(y+2)$。 答案 ... 已知 $\dfrac{1}{p}-\dfrac{1}{2q}=\dfrac{1}{3r}$， 当 $q=-1$ 且 $r=2$ 时，求 $p$ 的值。 答案 $p$ = ... 变形公式，使 $q$ 成为主项。 答案 $q$ = ... 解方程 $\frac{5}{x+2}-\frac{3x}{2x-1}=3. 将你的解精确到 2 位小数。 答案 x = ... 或 ...

1. Expand the bracket and simplify the equation.

$2(x-5)=3x-1$

$2x-10=3x-1$

2. Collect like terms and solve for $x$.

$2x-10=3x-1$

$-10=x-1$

$-9=x$

$x=-9$

1. Expand the bracket and collect like terms.

$7-4y>3(y+2)$

$7-4y>3y+6$

$7-6>3y+4y$

2. Solve for $y$ and write the inequality answer.

$1>7y$

$\dfrac{1}{7}>y$

Hence, $y<\dfrac{1}{7}$.

1. Substitute $q=-1$ and $r=2$ into the given formula.

Given
$\frac{1}{p}-\frac{1}{2q}=\frac{1}{3r}$
Substitute $q=-1$ and $r=2$:
$\frac{1}{p}-\frac{1}{2(-1)}=\frac{1}{3(2)}$
$\frac{1}{p}+\frac{1}{2}=\frac{1}{6}$

2. Solve the resulting equation to find $p$.

$\frac{1}{p}=\frac{1}{6}-\frac{1}{2}$
$\frac{1}{p}=\frac{1}{6}-\frac{3}{6}=-\frac{2}{6}=-\frac{1}{3}$
Hence
$p=-3$

1. Start from the given formula and clear the fractions.

$\dfrac{1}{p}-\dfrac{1}{2q}=\dfrac{1}{3r}$

Multiply throughout by $6pqr$:

$6qr-3pr=2pq$

2. Collect all terms involving $q$ on one side.

$6qr-3pr=2pq$

$6qr-2pq=3pr$

3. Factorise $q$ and divide to make $q$ the subject.

$6qr-2pq=3pr$

$q(6r-2p)=3pr$

$q=\dfrac{3pr}{6r-2p}=\dfrac{3pr}{2(3r-p)}$

1. State the restrictions and multiply by the common denominator.

$x\ne -2$ and $x\ne \dfrac12$.

Multiply both sides by $(x+2)(2x-1)$:
$\frac{5}{x+2}-\frac{3x}{2x-1}=3$
$5(2x-1)-3x(x+2)=3(x+2)(2x-1).$

2. Expand both sides and simplify to a quadratic equation.

$10x-5-3x^2-6x=3(2x^2+3x-2)$
$-3x^2+4x-5=6x^2+9x-6$
$9x^2+5x-1=0.$

3. Solve the quadratic equation.

For $9x^2+5x-1=0$,
$x=\frac{-5\pm\sqrt{5^2-4(9)(-1)}}{2(9)}$
$x=\frac{-5\pm\sqrt{61}}{18}.$

4. Calculate and round the two solutions to 2 decimal places.

$x=\frac{-5+\sqrt{61}}{18}=0.156\ldots \approx 0.16$
$x=\frac{-5-\sqrt{61}}{18}=-0.711\ldots \approx -0.71$
Both values are allowed since neither is $-2$ or $\dfrac12$.