Respuesta :
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
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
[tex] \left \{ {{6x-2y=4\:(I)} \atop {7x+5y=3\:(II)}} \right. [/tex]
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.
[tex]\left \{ {{6x-2y=4\:*(5)} \atop {7x+5y=3\:*(2)}} \right. [/tex]
[tex]\left \{ {{30x-10y=20\:} \atop {14x+10y=6\:}} \right. [/tex]
cancel (-10y and 10y)
[tex]\left \{ {{30x-\diagup\!\!\!\!\!10y=20\:} \atop {14x+\diagup\!\!\!\!\!10y=6\:}} \right. [/tex]
[tex] \left \{ {{30x=20} \atop {14x=6}} [/tex]
-----------------
44x = 26
[tex]x = \frac{26}{44}\:simplify\: \frac{\div2}{\div2} \to\: \boxed{x= \frac{13}{22} }[/tex]
Let's replace the value found in the second equation:
[tex]7x + 5y = 3 [/tex]
[tex]7* \frac{13}{22} + 5y = 3[/tex]
[tex]\frac{91}{22} + 5y = 3[/tex]
least common multiple (22)
[tex]\frac{91}{\diagup\!\!\!\!\!22} + \frac{110y}{\diagup\!\!\!\!\!22} = \frac{66}{\diagup\!\!\!\!\!22} [/tex]
[tex]91 + 110y = 66[/tex]
[tex]110y = 66 - 91[/tex]
[tex]110y = -25[/tex]
[tex]y = -\frac{25}{110}\:simplify\: \frac{\div5}{\div5} \to\:\boxed{y = -\frac{5}{22}} [/tex]
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.
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.
[tex]\left \{ {{6x-2y=4\:*(5)} \atop {7x+5y=3\:*(2)}} \right. [/tex]
[tex]\left \{ {{30x-10y=20\:} \atop {14x+10y=6\:}} \right. [/tex]
cancel (-10y and 10y)
[tex]\left \{ {{30x-\diagup\!\!\!\!\!10y=20\:} \atop {14x+\diagup\!\!\!\!\!10y=6\:}} \right. [/tex]
[tex] \left \{ {{30x=20} \atop {14x=6}} [/tex]
-----------------
44x = 26
[tex]x = \frac{26}{44}\:simplify\: \frac{\div2}{\div2} \to\: \boxed{x= \frac{13}{22} }[/tex]
Let's replace the value found in the second equation:
[tex]7x + 5y = 3 [/tex]
[tex]7* \frac{13}{22} + 5y = 3[/tex]
[tex]\frac{91}{22} + 5y = 3[/tex]
least common multiple (22)
[tex]\frac{91}{\diagup\!\!\!\!\!22} + \frac{110y}{\diagup\!\!\!\!\!22} = \frac{66}{\diagup\!\!\!\!\!22} [/tex]
[tex]91 + 110y = 66[/tex]
[tex]110y = 66 - 91[/tex]
[tex]110y = -25[/tex]
[tex]y = -\frac{25}{110}\:simplify\: \frac{\div5}{\div5} \to\:\boxed{y = -\frac{5}{22}} [/tex]
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.