Suggested: x-y+2z=7 3x+4y-5z=-5 2x-y+3z=12 - 3x+2y=5 2x-3y=7 class 10 - x-3y-7=0 3x-3y-15=0 - solve by cramer's rule 5x-7y+z=11 6x-8y-z=15 3x+2y-6z=7 - x+7y=10 3x-2y=7 - show that the straight lines 3x-5y+7=0 and 15x+9y+4=0 are perpendicular - 3x-y+7/11+2=10 and 2y+x+11/7=10 - increase by 7 the roots of the equation 3x^4+7x^3-15x^2+x-2=0 - 86- 15x-7(6x-9)-2 10x-5(2-3x) - x+y=7 3x-2y=11 - 3(15-4x)+5(3x-7)=15 - 34x-22x3-12x4-10x2-75 by 3x+7 - 4x+1/3+2x-1/2-3x-7/5=6 - 5x-2(2x-7)=2(3x-1)+7/2 solve - 1)(3x+7) Browse related:
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