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Use the substitution method to solve the following system of equations. x+2y-z=2 3x+y-2z=-15 3x+4y-3z=-2 Im lost!
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first equation : x = 2 -2y + z plug it into the second : 3(2-2y+z) + y -2z = -15 6 - 6y + 3z +y -2z = -15 z -5y = -21 z = -21 + 5y now x become : x= 2 -2y + (-21 + 5y) x = 2-2y-21+5y x = 3y - 19 now plug z,x into the third equation : 3(3y-19) +4y - 3(5y-21) = -2 9y -57 +4y -15y + 63 = -2 -2y = -8 y = 4 now plug y into x and z equations : x = 3*4 - 19 = -7 z = -21 +5*4 = -1 (-7,4,-1)
I got to -5y+z=-21 and I didnt know where to go from there. Thank You!
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