Practice

1. Solve for $x$

$$5-(x+7)+4x=2x-8$$

$$5 – (x+7) + 4x = 2x – 8$$

2. Solve for $x$ in terms of the other variables. State any restrictions.

$$a(x-b)=cx+d$$

$$a(x-b) = cx + d$$

3. Solve for $x$ in terms of the other variables. State any restrictions.

$$(a+b)x+c=dx+ex+f$$

$$(a+b)x + c = dx + ex + f$$

4. Solve for $x$ in terms of the other variables. State any restrictions.

$$a[b+c(x-d)]+e(fx+g)=h[i+jx]+kx+l$$

$$a[b+c(x-d)] + e(fx + g) = h[i + jx] + kx + l$$

5. Solve for $x$ in terms of the other variables. State any restrictions.

$$\frac{x-a}{b}+\frac{x}{c}=d$$

$$\frac{x-a}{b} + \frac{x}{c} = d$$

6. Solve for $x$ in terms of the other variables. State any restrictions.

$$p=\frac{ax+b}{cx+d}$$

$$p \;=\; \frac{ax + b}{cx + d}$$

7. Solve for $x$ in terms of the other variables.

$$k(x+a) \;=\; (x+b)^2 – (x+c)^2$$

Solutions

1)

$$5 – (x+7) + 4x = 2x – 8$$ $$5 – x – 7 + 4x = 2x – 8$$ $$-2 + 3x = 2x – 8$$ $$-2 + x = -8$$ $$x = -6$$

2)

$$a(x-b)=cx+d$$ $$ax – ab = cx + d$$ $$ax – cx = ab + d$$ $$x(a-c) = ab + d$$ $$x = \frac{ab + d}{a – c},\quad \text{provided } a \ne c.$$

3)

$$(a+b)x + c = dx + ex + f$$ $$ax + bx + c = dx + ex + f$$ $$ax + bx – dx – ex = f – c$$ $$x(a + b – d – e) = f – c$$ $$x = \frac{f – c}{a + b – d – e},\quad \text{provided } a + b – d – e \ne 0.$$

4)

$$a[b+c(x-d)] + e(fx + g) = h[i + jx] + kx + l$$ $$ab + ac(x-d) + efx + eg = hi + hjx + kx + l$$ $$ab + acx – acd + efx + eg = hi + hjx + kx + l$$ $$acx + efx – hjx – kx = hi + l – eg – ab + acd$$ $$x(ac + ef – hj – k) = hi + l – eg – ab + acd$$ $$x = \frac{hi + l – eg – ab + acd}{ac + ef – hj – k},\quad \text{provided } ac + ef – hj – k \ne 0.$$

5)

$$\frac{x-a}{b} + \frac{x}{c} = d$$

Multiply by $bc$ (restrictions: $b\ne 0,\ c\ne 0$):

$$c(x-a) + bx = dbc$$ $$cx – ca + bx = dbc$$ $$(b + c)x = dbc + ca$$ $$x = \frac{dbc + ca}{b + c},\quad \text{provided } b + c \ne 0.$$

6)

$$p = \frac{ax + b}{cx + d}$$

Cross-multiply (restriction: $cx + d \ne 0$):

$$p(cx + d) = ax + b$$ $$pcx + pd = ax + b$$ $$(pc – a)x = b – pd$$ $$x = \frac{b – pd}{pc – a},\quad \text{provided } pc – a \ne 0 \text{ and } cx + d \ne 0 \text{ for this } x.$$

7)

$$k(x+a) = (x+b)^2 – (x+c)^2$$

Use difference of squares:
$(x+b)^2 – (x+c)^2 = \big[(x+b)-(x+c)\big]\big[(x+b)+(x+c)\big] = (b – c)(2x + b + c).$
So

$$kx + ka = 2(b – c)x + (b – c)(b + c)$$ $$\big(k – 2(b – c)\big)x = (b – c)(b + c) – ka$$ $$x = \frac{(b – c)(b + c) – ka}{\,k – 2(b – c)\,},\quad \text{provided } k – 2(b – c) \ne 0.$$