Question

ws 12 p.3
lesson 3 - adding and subtracting polynomials unit 3 - algebraic expressions
mth11w
jensen ws 12 p.1
part 1: adding polynomials
polynomial: an algebraic expression consisting of one or more terms connected by addition or
subtraction operators.
when adding polynomials, you can simply drop the brackets and collect the like terms.
for example:
(4x + 3) + (7x + 2)
= 4x + 3 + 7x + 2
step 1: remove the brackets
step 2: rearrange like terms into groups
step 3: collect the like terms
example 1: simplify the following expressions
a) (3y + 5) + (7y - 4) b) (2p - 2) + (4p - 7)
c) (6x - 12) + (-9x - 4) + (x + 14) d) (5x - 4y - 1) + (-2x + 5y + 13)
= 5x - 4y - 1 + -2x + 5y + 13
= 3x - 4y - 1 + 5y + 13
= 3x + y - 1 + 13
= 3x + y + 12

Explanation:

Part 1: Solving each sub - question (a, b, c, d)

a) Simplify \((3y + 5)+(7y - 4)\)

Step 1: Remove the Brackets

Since there is a plus sign in front of each bracket, we can directly remove the brackets. So we get \(3y + 5+7y - 4\).

Step 2: Rearrange like terms into groups

Group the \(y\) - terms and the constant terms together. We have \((3y + 7y)+(5 - 4)\).

Step 3: Collect the like terms

For the \(y\) - terms: \(3y+7y = 10y\). For the constant terms: \(5 - 4=1\). So the simplified expression is \(10y + 1\).

b) Simplify \((2p-2)+(4p - 7)\)

Step 1: Remove the Brackets

Because of the plus sign in front of the brackets, we remove the brackets to get \(2p-2 + 4p-7\).

Step 2: Rearrange like terms into groups

Group the \(p\) - terms and the constant terms: \((2p + 4p)+(-2-7)\).

Step 3: Collect the like terms

For the \(p\) - terms: \(2p + 4p=6p\). For the constant terms: \(-2-7=-9\). So the simplified expression is \(6p-9\).

c) Simplify \((6x - 12)+(-9x - 4)+(x + 14)\)

Step 1: Remove the Brackets

When we remove the brackets, we have \(6x - 12-9x - 4+x + 14\).

Step 2: Rearrange like terms into groups

Group the \(x\) - terms and the constant terms: \((6x-9x + x)+(-12-4 + 14)\).

Step 3: Collect the like terms

For the \(x\) - terms: \(6x-9x+x=(6 - 9 + 1)x=-2x\). For the constant terms: \(-12-4 + 14=(-16)+14=-2\). Wait, let's recalculate the constant terms: \(-12-4=-16\), \(-16 + 14=-2\)? Wait, no: \(-12-4+14=(-12 + 14)-4=2 - 4=-2\)? Wait, no, \(6x-9x+x=(6 + 1-9)x=(-2)x\), and \(-12-4 + 14=(-16)+14=-2\)? Wait, no, \(-12-4=-16\), \(-16 + 14=-2\)? Wait, actually, \(-12-4+14=(-12)+(-4)+14=-16 + 14=-2\)? Wait, but let's do it again: \(6x-9x+x=(6 - 9+1)x=-2x\). \(-12-4 + 14=-16 + 14=-2\)? Wait, that can't be right. Wait, \(-12-4=-16\), \(-16+14=-2\). But let's check: \(6x-9x+x=(6x+x)-9x=7x-9x=-2x\). \(-12-4 + 14=(-12 + 14)-4=2-4=-2\). So the simplified expression is \(-2x-2\)? Wait, no, wait: \(-12-4+14=-12+( - 4)+14=(-12 + 14)+(-4)=2-4=-2\). Yes. Wait, but maybe I made a mistake. Let's substitute \(x = 1\) into the original expression: \((6(1)-12)+(-9(1)-4)+(1 + 14)=(6 - 12)+(-9 - 4)+15=(-6)+(-13)+15=-4\). Now substitute \(x = 1\) into \(-2x-2\): \(-2(1)-2=-4\). Oh, right. So the simplified expression is \(-2x-2\)? Wait, no, wait: \(6x-9x+x=(6 - 9 + 1)x=-2x\). \(-12-4 + 14=-16 + 14=-2\). So \(-2x-2\).

d) We already have the hand - written solution, but let's re - do it properly. Simplify \((5x-4y - 1)+(-2x + 5y+13)\)

Answer:

Step 1: Remove the Brackets

We get \(5x-4y - 1-2x + 5y+13\).

Step 2: Rearrange like terms into groups

Group the \(x\) - terms, \(y\) - terms and the constant terms: \((5x-2x)+(-4y + 5y)+(-1 + 13)\).

Step 3: Collect the like terms

For the \(x\) - terms: \(5x-2x = 3x\). For the \(y\) - terms: \(-4y+5y=y\). For the constant terms: \(-1 + 13 = 12\). So the simplified expression is \(3x + y+12\).

Final Answers:

a) \(\boldsymbol{10y + 1}\)

b) \(\boldsymbol{6p-9}\)

c) \(\boldsymbol{-2x-2}\)

d) \(\boldsymbol{3x + y+12}\)