Equations
When simplifying algebraic expressions, we perform operations within parentheses first and then exponents and then multiplication and then division and then addition and lastly subtraction. This can be remembered by the mnemonic:PEMDAS
Please Excuse My Dear Aunt Sally
When solving equations, however, we apply the mnemonic in reverse order: SADMEP. This is often expressed as follows: inverse operations in inverse order. The goal in solving an equation is to isolate the variable on one side of the equal sign (usually the left side). This is done by identifying the main operation—addition, multiplication, etc.—and then performing the opposite operation.
Solve the following equation for x: 2x + y = 5
Solution: The main operation is addition (remember addition now comes before multiplication, SADMEP), so subtracting y from both sides yields
2x + y – y = 5 – y
Simplifying yields
2x = 5 – y
The only operation remaining on the left side is multiplication. Undoing the multiplication by dividing both sides by 2 yields
Canceling the 2 on the left side yields
Solve the following equation for x: 3x – 4 = 2(x – 5)
Solution: Here x appears on both sides of the equal sign, so let’s move the x on the right side to the left side. But the x is trapped inside the parentheses. To release it, distribute the 2:
3x – 4 = 2x – 10
Now, subtracting 2x from both sides yields
x – 4 = –10
Finally, adding 4 to both sides yields
x = –6
Equation |
Deduction |
y – x = 1 |
y > x |
y^{2} = x^{2} |
y = ± x, or . That is, x and y can differ only in sign. |
y^{3} = x^{3} |
y = x |
y = x^{2} |
y ≥ 0 |
y > 0 | |
Both x and y are positive or both x and y are negative. | |
x^{2} + y^{2} = 0 |
y = x = 0 |
3y = 4x and x > 0 |
y > x and y is positive. |
3y = 4x and x < 0 |
y < x and y is negative. |
y ≥ 0 and x ≥ –2 | |
y = 2x |
y is even |
y = 2x + 1 |
y is odd |
yx = 0 |
y = 0, or x = 0, or both |
If a + 3a is 4 less than b + 3b, then a – b =
A. –4
B. –1
C. 1/5
D. 1/3
E. 2
Translating the sentence into an equation gives a + 3a = b + 3b – 4
Combining like terms gives 4a = 4b – 4
Subtracting 4b from both sides gives 4a – 4b = –4
Finally, dividing by 4 gives a – b = –1
Hence, the answer is (B).
If w ≠ 0 and , what is the value of w – x in terms of y ?
A. 2y
B.
C.
D.
E. y
The equation stands for three equations: w = 2x, , and . From the last equation, we get ; and from the second equation, we get . Hence,
Hence, the answer is (B).
If p and q are positive, p^{2} + q^{2} = 16, and p^{2} – q^{2} = 8, then q =
A 2
B 4
C 8
D
E
Subtract the second equation from the first: p^{2} + q^{2} = 16
(–) p^{2} – q^{2} = 8
2q^{2} = 8
Dividing both sides of the equation by 2 gives q^{2} = 4
Finally, taking the square root of both sides gives q = ±2
Hence, the answer is (A).