SOLVING LITERAL EQUATIONS ON SAT

Problem 1 :

If X = (X + 1)/(Y + Z), find X in terms of Y and Z.

Solution :

X = (X + 1)/(Y + Z)

Multiply (Y + Z) on both sides.

X(Y + Z) = X + 1

XY + XZ = X + 1

XY + XZ – X = 1

X(Y + Z - 1) = 1

X = 1/(Y + Z -1)

Problem 2 :

If x(y + 2) = y, find y in terms of x.

Solution :

x(y + 2) = y

xy + 2x = y

2x = y – xy

2x = y(1 – x)

2x/(1 – x) = y

Problem 3 :

If a/b = (a + 1)/2c, find a in terms of b and c.

Solution :

a/b = (a + 1)/2c

Using cross – multiply,

2ac = (a + 1)b

2ac = ab + b

2ac – ab = b

a(2c – b) = b

a = b/(2c – b)

Problem 4 :

If t = (2/3)ax, find ax in terms of t.

Solution :

t = (2/3)ax

To find ax :

Using cross – multiply,

3t/2 = ax

Problem 5 :

If 3x + 6y = 7z, find x + 2y in terms of z.

Solution :

3x + 6y = 7z

To find x + 2y :

3(x + 2y) = 7z

Dividing 3 on each sides.

x + 2y = 7z/3

Problem 6 :

If x + 5 = 2b, find 2x + 10 in terms of b.

Solution :

x + 5 = 2b

To find 2x + 10 :

Multiply 2 on both sides.

2(x + 5) = 2(2b)

2x + 10 = 4b

Problem 7 :

If (a – 1)/2t = a, find 4t in terms of a.

Solution :

(a – 1)/2t = a

To find 4t :

Using cross – multiplication.

(a – 1)/a = 2t

Multiply 2 on each sides.

2(a – 1)/a = 2(2t)

(2a – 2)/a = 4t

2(a – 1)/a = 4t

Problem 8 :

If (p – h)/(p + h) = 2/3, find p/h.

Solution :

(p – h)/(p + h) = 2/3

To find p/h :

Using cross – multiplication,

3(p - h) = 2(p + h)

3p – 3h = 2p + 2h

3p – 2p = 2h + 3h

p = 5h

p/h = 5

Problem 9 :

If (1 + 2r)/(1 – t) = 1/2, find t in terms of r.

Solution :

(1 + 2r)/(1 – t) = 1/2

To find t :

Using cross – multiplication,

2(1 + 2r) = 1 – t

2 + 4r = 1 – t

2 – 1 + 4r = -t

1 + 4r = -t

t = -1 – 4r

Problem 10 :

If x^y = z, then find x^2y in terms of z.

Solution :

x^y = z

To find x^2y :

Squaring on both sides.

(x^y)² = z²

x^2y = z²

Problem 11 :

If (4^{x + 1})/(x³ – x²) = p(x⁵ – x⁴), what is p in terms of x ?

Solution :

(4^{x + 1})/(x³ – x²) = p(x⁵ – x⁴)

Dividing (x⁵ – x⁴) on each sides.

(4^{x + 1})/(x³ – x²) × (x⁵ – x⁴) = p(x⁵ – x⁴)/(x⁵ – x⁴)

(4^{x + 1})/(x³ – x²) × (x⁵ – x⁴) = p

p = (4^{x + 1})/x²(x³ – x²)²

Problem 12 :

If 2^x(x³ – 1/x) = m(x² + 1) – 1/x², what is m in terms of x ?

Solution :

2^x(x³ – 1/x) = m(x² + 1) – 1/x²

2^x(x³ – 1/x) + 1/x² = m(x² + 1)

Dividing (x² + 1) on each sides.

m = (2^x(x³ – 1/x) + 1/x²)/(x² + 1)

Problem 13 :

If ((√x + 1)/(5x² – 3) – x³) = 1/nx, what is n in terms of x ?

Solution :

((√x + 1)/(5x² – 3) – x³) = 1/nx

n((√x + 1)/(5x² – 3) – x³) = 1/x

n = 1/(x((√x + 1)/(5x² – 3) – x³))

Problem 14 :

If a(b² + 2) + c = 5(c + 1)³, what is a in terms of b and c ?

Solution :

a(b² + 2) + c = 5(c + 1)³

a(b² + 2) = 5(c + 1)³ – c

Dividing (b² + 2) on each sides.

a(b² + 2)/(b² + 2) = (5(c + 1)³ – c)/(b² + 2)

a = (5(c + 1)³ – c)/(b² + 2)

Problem 15 :

If k(x² + 4) + ky = (7x² + 3)/2, what is k in terms x and y ?

Solution :

k(x² + 4) + ky = (7x² + 3)/2

k(x² + 4 + y) = (7x² + 3)/2

Dividing (x² + 4 + y) on each sides.

k = (7x² + 3)/(2(x² + 4 + y))

Problem 16 :

If ax + 3a + x + 3 = b, what is x in terms a and b ?

Solution :

ax + 3a + x + 3 = b

ax + x = b - 3a - 3

x(a + 1) = b - 3a - 3

x = (b - 3a - 3)/(a + 1)