Question

A system of equations is given below.
2x + 7y = 1
-3x – 4y = 5

Create an equivalent system of equations by replacing the first equation by multiplying the first equation by an integer other than 1, and adding it to the second equation.
Use any method to solve the equivalent system of equations (the new first equation with the original second equation).
Prove that the solution for the equivalent system is the same as the solution for the original system of equations.

Answer 1

So, I already did some of part a, multiplied the first equation by 2, and got 4x+14y=2. What should I do next?

Answer 2

Hmm, that wording is confusing to me

Answer 3

ikr

Answer 4

Plz help!

Answer 5

multiply the first one by 3 and the second one by 21 (all the way across)
then when you add them the x terms will go bye bye

Answer 6

ok not "21"
multiply the second one by 2

Answer 7

Thanks! But then what do you do? @satellite73

Answer 8

supposed to do 2*r1 + r2 => r1

Answer 9

then solve system comprised of new r1 and original r2.

Answer 10

add

Answer 11

in a column

Answer 12

the first term of the first equation will be \(6x\) and the first term of the second one will be \(-6x\)
when you add the two together, there will be no more \(x\)

Answer 13

\[2x + 7y = 1 \\-3x – 4y = 5\]\[6x+21y=3\\
-6x-8y=10\] now add them up

Answer 14

So it would be 13y=-7?

Answer 15

@satellite73

Answer 16

After you multiplied the first equation by two and got you forgot to add the two to create a third equation, correct?

eq 1) 2x + 7y = 1 times 2

eq 3) 4x+14y=2 add eq 3 to eq 2
eq 2) -3x – 4y = 5

eq 4) x + 10y = 7

then it says to use the FIRST new equation ( eq 3) to solve the equivalent system with the original eq 2. Why did I have to add eq 2 and 3 together? Or do they want me to solve the equivalent system using eq 2 and 4 ? Which may or may not yield the correct answer.

Answer 17

Ohhh I think I understand now. Thanks @retirEEd Do you maybe know how to do part c?

Answer 18

solve new system of eqn 4 and eqn 2 and show the soultion is the same as the solution of the original system

Answer 19

Thanks pgpilot326, I'll try it. I got the solution for eqn 1 and eqn 2 and I like your short for equation better!

Answer 20

Yes it works! eqn 1 and eqn 2 solve to be x=-3 and y= 1 as does eqn 2 and eqn 4.

Much easier to visualize on a graphing calculator. Three lines all intersection at (-3,1), then typing all of this out.

eqn 1) 2x + 7y = 1
x = ((1-7y)/2
Plug x into eqn 2
eqn 2) -3x – 4y = 5
-3[(1-7y)/2] - 4y = 5
-3/2 + 21y/2 - 8y/2 = 10/2

13y/2 = 10/2 +3/2 multiply both sides by 2 to clear the fraction
13y = 13 y = 1

plug y =1 into eqn 1
2x + 7(1) = 1
x = (1-7)/2 = -6/2 x= -3

eqn 4) x + 10y = 7
eqn 2) -3x – 4y = 5

multiply eqn 4 by 3 and add to eqn 2 the x term equates to zero and

30y + (-4y) = 21 + 5 26y = 26 y = 1 as before

plug y = 1 into eqn 4
x + 10(1) = 7 x = 7 -10 x = -3 as before