s.gracex
  • s.gracex
Which is the solution to the system of equations? Use the linear combination method (with or without multiplication) 4x-5y=-18 3x+2y=-2 A. (3,-1) B. (-1,3) C. (-2,7) d. (-2,2)
Mathematics
  • Stacey Warren - Expert brainly.com
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SOLVED
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jamiebookeater
  • jamiebookeater
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whpalmer4
  • whpalmer4
linear combination means you add the equations together (possibly multiplying one or both by a constant first) so that they combine and make one equation in one variable instead of two equations in two variables. For this to work, you need to have the coefficients for one of the variables to be equal in magnitude but opposite in sign. Do you have that situation here?
s.gracex
  • s.gracex
no?
whpalmer4
  • whpalmer4
That's right, you do not. So, how can we get that to be the case?

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s.gracex
  • s.gracex
i'm not sure...
whpalmer4
  • whpalmer4
\[4x-5y=-18\]\[ 3x+2y=-2 \] What if we multiply the first equation by 2, and the second equation by 5? I'm looking at the \(y\) column and multiplying each equation by the coefficient of \(y\) in the other equation. That will give us: \[2*4x -2*5y = 2*(-18)\]\[5*3x+5*2y = 5*(-2)\]simplifying both, we get: \[8x-10y = -36\]\[15x+10y=-10\] Now we have two equations where if we add them together, one set of variables will combine to 0, leaving us with an equation in only 1 variable. \[8x-10y = -36\]\[15x+10y=-10\] -------------------------- \[8x+15x -10y + 10y = -36-10\]\[23x = -46\] Now you solve that equation for \(x\). When you have the value of \(x\), plug it into any of the equations and find the value of \(y\). What do you get?

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