Solve $2x-7=3(1-3x).\n\nAnswer x = ... Solve the inequalities $7x-1<13$ and $\frac{x+1}{2}\ge -2(x-2).\n\nAnswer ... Rearrange the formula $y=\frac{x^2+5}{7x^2}$ to make x the subject.\n\nAnswer x = ... Solve the equation $\frac{2x}{1-x} - 3 = \frac{1}{2x-3}$. Give your solutions correct to two decimal places. Answer x = ... or ...

1. Expand the bracket on the right-hand side.

$2x-7=3(1-3x)$

$2x-7=3-9x$

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

$2x+9x=3+7$

$11x=10$

$x=\dfrac{10}{11}$

1. Solve the first inequality.

$7x-1<13$

$7x<14$

$x<2$

2. Solve the second inequality.

$\dfrac{x+1}{2}\ge -2(x-2)$

Multiply both sides by $2$:

$x+1\ge -4(x-2)$

$x+1\ge -4x+8$

$5x\ge 7$

$x\ge \dfrac{7}{5}$

3. Combine the two inequalities to give the common solution range.

Since both inequalities must be true,

$x<2$ and $x\ge \dfrac{7}{5}$.

Therefore,

$\dfrac{7}{5}\leq x<2$.

1. Clear the fraction by multiplying both sides by $7x^2$.

$y=\dfrac{x^2+5}{7x^2}$

$7x^2y=x^2+5$

2. Collect the $x^2$ terms on one side and factorise.

$7x^2y-x^2=5$

$x^2(7y-1)=5$

3. Divide by the coefficient of $x^2$ and take the square root.

$x^2=\dfrac{5}{7y-1}$

$x=\pm\sqrt{\dfrac{5}{7y-1}}$

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

The denominators must not be zero, so $x\ne 1$ and $x\ne \dfrac{3}{2}$.

$\frac{2x}{1-x}-3=\frac{1}{2x-3}$

Multiply throughout by $(1-x)(2x-3)$:

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

2. Expand and simplify to form a quadratic equation.

$4x^2-6x-3(1-x)(2x-3)=1-x$

Since
$(1-x)(2x-3)=-2x^2+5x-3,$

$4x^2-6x-3(-2x^2+5x-3)=1-x$

$4x^2-6x+6x^2-15x+9=1-x$

$10x^2-21x+9=1-x$

$10x^2-20x+8=0$

Divide by $2$:

$5x^2-10x+4=0$

3. Solve the quadratic equation.

For $5x^2-10x+4=0$,

$x=\frac{-(-10)\pm\sqrt{(-10)^2-4(5)(4)}}{2(5)}$

$x=\frac{10\pm\sqrt{100-80}}{10}$

$x=\frac{10\pm\sqrt{20}}{10}$

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

$x=\frac{10+\sqrt{20}}{10}=1.447\ldots \approx 1.45$

$x=\frac{10-\sqrt{20}}{10}=0.552\ldots \approx 0.55$

Both are allowed since neither is $1$ nor $\dfrac{3}{2}$.