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5 votes
What would you need to multiply by in order to solve this system of equations using the linear combination (elimination) method?

6x-2y=4
7x+5y=3
Question 9 options:

Multiply the 1st equation by 3 and leave the 2nd equation alone. This will eliminate the y's when you add the two new equations together.


Multiply the 2nd equation by -6 and leave the 1st equation alone. This will eliminate the x's when you add the two new equations together.


Multiply the 1st equation by 7 and multiply the 2nd equation by 4. This will eliminate the x's when you add the two new equations together.


Multiply the 1st equation by 5 and multiply the 2nd equation by 2. This will eliminate the y's when you add the two new equations together.

2 Answers

3 votes

\left \{ {{6x-2y=4\:(I)} \atop {7x+5y=3\:(II)}} \right.
Multiply the 1st equation by 5 and multiply the 2nd equation by 2. This will eliminate the y's when you add the two new equations together.

\left \{ {{6x-2y=4\:*(5)} \atop {7x+5y=3\:*(2)}} \right.

\left \{ {{30x-10y=20\:} \atop {14x+10y=6\:}} \right.
cancel (-10y and 10y)

\left \{ {{30x-\diagup\!\!\!\!\!10y=20\:} \atop {14x+\diagup\!\!\!\!\!10y=6\:}} \right.

\left \{ {{30x=20} \atop {14x=6}}
-----------------
44x = 26

x = (26)/(44)\:simplify\: (/2)/(/2) \to\: \boxed{x= (13)/(22) }

Let's replace the value found in the second equation:

7x + 5y = 3

7* (13)/(22) + 5y = 3

(91)/(22) + 5y = 3
least common multiple (22)

(91)/(\diagup\!\!\!\!\!22) + (110y)/(\diagup\!\!\!\!\!22) = (66)/(\diagup\!\!\!\!\!22)

91 + 110y = 66

110y = 66 - 91

110y = -25

y = -(25)/(110)\:simplify\: (/5)/(/5) \to\:\boxed{y = -(5)/(22)}

Answer:
To find the roots, use the last option of the statement.

Multiply the 1st equation by 5 and multiply the 2nd equation by 2. This will eliminate the y's when you add the two new equations together.





User AMunim
by
8.4k points
0 votes
let's see the least common multiplules of the x and y terms

x terms
6 and 7, least common multipule is 42 so you would have to multiply the first euqaiton by -7 and 2nd equation by 6, but that isn't an option
therefor we aren't supposed to eliminate the x terms


y terms
-2 and 5
LCM=10 so multiply first by 5 and 2nd by 2
that is last option
User Gustavo Soler
by
8.8k points

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