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solve for x₁ and x₂
6x₁+3x₂=60
6x₁-2x₂=80

Sagot :

secretlyblu
first let's change the "x1" to "A" Nd the "x2" to "B" __ so now the problem reads "6A + 3B = 60" And "6A - 2B= 80" we want to isolate variable. and the easiest one is the "B" by getting rid of the "A". set up a subtraction problem where you take one equation away from the other. 6a-6a=0. 3b - (-2b) = 5b and 60-80=-20. so your new equation reads 5B= -20. you wanna get B on one side so yu divide 5 on both sides giving you b= -4 or "x2"=-4. from there you can plug -4 into any one of the original equations. ex: 6A -2(-4)=80 then 6A + 8=80 then you subtract 8 from both sides because you're trying to isolate the "A" variable. which leaves you with 6A = 72. and now you wanna get"A" on one side so yu divide 6 from both sides and get A=12 or "x1"=12. do a quick check and plug in these numbers into any one of the equation.
kate200468
[tex] \left \{ {\big{6x_1+3x_2=60\ \ \ \ \ \ \ \ \ \ \ } \atop \big{6x_1-2x_2=80\ /\cdot(-1)}} \right. \\\\ \left \{ {\big{6x_1+3x_2=60\ \ \ \ } \atop \big{-6x_1+2x_2=-80}} \right. \\-----------\\3x_2+2x_2=60-80\\5x_2=-20\ /:5\\\\x_2=-4\\\\6x_1+3x_2=60\ /:3\\2x_1+x_2=20\ \ \ \Rightarrow\ \ \ 2x_1+(-4)=20\ \ \ \Rightarrow\ \ \ 2x_1=24\ /:2\\\\x_1=12\\\\ \left \{ {\big{x+1=12} \atop\big {x_2=-4}} \right. [/tex]
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