Question

Need help here please....

7x+2y= 7/4
y=4x-1
The solution is ? (Type an ordered pair.)

Answer 1

you could replace y in the first equation with 4x-1

Answer 2

7x + 2y = 7/4

y = 4x - 1

substitute second eqn in the first eqn

7x + 2 (4x-1) = 7/4

solve for x

Answer 3

Can you help me break it down, I am not great with fractions...
(4x-1)= 3x?

Answer 4

7x + 2 (4x-1) = 7/4
start with chan's equation:
distribute the 2 over (4x-1) by multiplying everything inside the parens by 2

Answer 5

8x-2=6x (that's what I got when I multiplied everything inside by 2)

Answer 6

first: 2(4x-1) is 8x-2 which is what you got.
next, you should get used to writing down the whole equation. so you would get
7x+2(4x-1)= 7/4
7x+8x-2 =7/4

now this is important: you can add 7x's to 8x's and get 15 x's
but 8x-2 cannot be replaced with 6x (8x-2x is 6x)

combine the x terms
add +2 to both sides

Answer 7

7x+2(4x-1)=7/4
7x+8x-2=7/4
15x-2=7/4

Answer 8

am I on the right track?

Answer 9

yes. now +2 on both sides

Answer 10

15x-2=7/4
17x-4=7/4

Answer 11

here is how:
15x-2=7/4
write +2 on both sides:
15x-2+2 = 7/4 +2
simplify the left side. and simplify the right side

Answer 12

you can add or subtract numbers, so -2+2 becomes 0

on the right side, use a common denominator of 4: multiply 2 (2/1) by 4/4
then add the fractions

Answer 13

15x-2+2=15x
7/4+2= 3 3/4

Answer 14

I got the first one, but I think the second part of 7/4 +2 is wrong

Answer 15

I would write it like this:
15x-2+2 = 7/4 +2
15x= 7/4 + 8/4
15x= 15/4 (add the tops after you have a common denominator)

do you know what to do next to find x?

Answer 16

7/4+8/4 =15/4
so, 15x=15/4
do I add 15 to both sides?

Answer 17

you could but it would not be helpful
what do you get if you multiplied both sides by 1/15:
\[ \frac{1}{15}\cdot \frac{15x}{1} = \frac{1}{15}\cdot \frac{15}{4}\]

Answer 18

15/60

Answer 19

well, I did 1*15 then 15*4
to arrive at that answer

Answer 20

yes. but you should write the entire equation

Answer 21

1/15* 15/4=15/60

Answer 22

the equation with x in it

Answer 23

1/15*15x/1 that side I have no idea

Answer 24

is it zero?

Answer 25

that is the left side of the equation. you should write
\[ \frac{1}{15}\cdot \frac{15x}{1} = \frac{15}{60} \]
(this way you won't get lost as you take each step)

But here is the "easy way":
when you multiply fractions, you multiply top times top and bottom by bottom
\[ \frac{1}{15}\cdot \frac{15x}{1} = \frac{1}{15}\cdot\frac{15}{4} \]
you get
\[ \frac{1\cdot 15 \cdot x}{15}= \frac{1\cdot15}{15 \cdot 4}\]
order of multiplication can change (right?) 1*15= 15*1
that means this is the same as
\[ \frac{15 \cdot x}{15}= \frac{15 \cdot 1}{15 \cdot 4}\]
or "undoing" the fractions
\[ \frac{15}{15} \cdot \frac{x}{1}= \frac{15 }{15}\cdot \frac{1}{4}\]
what is 15/15 ?

Answer 26

Once you learn this idea, you see that you can "cancel"
\[ \frac{1}{\cancel{15}}\cdot \frac{\cancel{15}x}{1} = \frac{1}{\cancel{15}}\cdot \frac{\cancel{15}}{4} \]