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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.

Verified Answer :

apologiabiology

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

dexteright02

\(\left \{ {{6x-2y=4\:(I)} \atop {7x+5y=3\:(II)}} \right.\)
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
\(x = \frac{26}{44}\:simplify\: \frac{\div2}{\div2} \to\: \boxed{x= \frac{13}{22} }\)

Let's replace the value found in the second equation:
\(7x + 5y = 3\)
\(7* \frac{13}{22} + 5y = 3\)
\(\frac{91}{22} + 5y = 3\)
least common multiple (22)
\(\frac{91}{\diagup\!\!\!\!\!22} + \frac{110y}{\diagup\!\!\!\!\!22} = \frac{66}{\diagup\!\!\!\!\!22}\)
\(91 + 110y = 66\)
\(110y = 66 - 91\)
\(110y = -25\)
\(y = -\frac{25}{110}\:simplify\: \frac{\div5}{\div5} \to\:\boxed{y = -\frac{5}{22}}\)

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.





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