Math  /  Algebra

QuestionDate:
Forms of lines Exit Ticket 1) [3 marks] Write the following in the form ax+by=ca x+b y=c where a,ba, b, and cc are integers. y=56x+3y=\frac{5}{6} x+3 2) [2 marks] Write the following in the form y=ax+by=a x+b. 4x+3y=14 x+3 y=-1 3) (3(3 marks/ Write the following in the form ax+by=ca x+b y=c where a,ba, b, and cc are integers. y5=4(x+3)y-5=-4(x+3) 4) [2[2 marks ]] Explain why the gradient of the line with the form ax+by=ca x+b y=c is ab-\frac{a}{b}.

Studdy Solution

STEP 1

What is this asking? We need to rewrite equations of lines in different forms, shuffling things around using algebra! Watch out! Don't forget those integer coefficients and watch your signs when rearranging terms.

STEP 2

1. Convert to ax + by = c
2. Convert to y = ax + b
3. Convert to ax + by = c, again!
4. Explain the gradient

STEP 3

Alright, we're given y=56x+3y = \frac{5}{6}x + 3.
We want it in the form ax+by=cax + by = c, where aa, bb, and cc are integers.

STEP 4

First, let's **get rid of that fraction**!
Multiply both sides of the equation by **6** to eliminate the denominator: 6y=6(56x+3)6 \cdot y = 6 \cdot \left(\frac{5}{6}x + 3\right) 6y=5x+186y = 5x + 18

STEP 5

Now, we want the x and y terms on the same side.
Let's **subtract** 5x5x from both sides: 6y5x=5x+185x6y - 5x = 5x + 18 - 5x 5x+6y=18-5x + 6y = 18Boom! We have our equation in the form ax+by=cax + by = c, where a=5a = \mathbf{-5}, b=6b = \mathbf{6}, and c=18c = \mathbf{18}.

STEP 6

We're given 4x+3y=14x + 3y = -1 and we want it in the form y=ax+by = ax + b.
This means we need to **isolate** yy.

STEP 7

Let's **subtract** 4x4x from both sides: 4x+3y4x=14x4x + 3y - 4x = -1 - 4x 3y=4x13y = -4x - 1

STEP 8

Now, **divide** both sides by **3**: 3y3=4x13\frac{3y}{3} = \frac{-4x - 1}{3} y=43x13y = -\frac{4}{3}x - \frac{1}{3}And there we have it!
Our equation is now in the form y=ax+by = ax + b, where a=43a = \mathbf{-\frac{4}{3}} and b=13b = \mathbf{-\frac{1}{3}}.

STEP 9

We have y5=4(x+3)y - 5 = -4(x + 3).
We want it in ax+by=cax + by = c form, with integer coefficients.

STEP 10

First, **distribute** the 4-4 on the right side: y5=4x12y - 5 = -4x - 12

STEP 11

Now, let's **add** 4x4x to both sides and **add** 55 to both sides to get those integer coefficients: y5+4x+5=4x12+4x+5y - 5 + 4x + 5 = -4x - 12 + 4x + 5 4x+y=74x + y = -7So, we have a=4a = \mathbf{4}, b=1b = \mathbf{1}, and c=7c = \mathbf{-7}.

STEP 12

We're asked to explain why the gradient of ax+by=cax + by = c is ab-\frac{a}{b}.
Remember, the gradient is the same as the slope.

STEP 13

Let's **rearrange** the equation ax+by=cax + by = c to solve for yy, just like we did before. **Subtract** axax from both sides: by=ax+cby = -ax + c

STEP 14

Now, **divide** both sides by bb: y=abx+cby = -\frac{a}{b}x + \frac{c}{b} This is now in slope-intercept form (y=mx+by = mx + b), where mm is the slope or gradient.
We can see clearly that the **coefficient** of xx is ab-\frac{a}{b}, which represents our gradient!

STEP 15

1. 5x+6y=18-5x + 6y = 18
2. y=43x13y = -\frac{4}{3}x - \frac{1}{3}
3. 4x+y=74x + y = -7
4. The gradient is ab-\frac{a}{b} because when we solve for yy in ax+by=cax + by = c, we get y=abx+cby = -\frac{a}{b}x + \frac{c}{b}, which is in slope-intercept form, revealing the gradient as the coefficient of xx.

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