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x+5y-4z=-10 equation 1 2x-y+5z=-9 equation 2 2x-10y-5z=0 equation 3 okay so you want to rearrange an equation lets rearrange the first one and make x the subject x+5y-4z=-10 equation 1 x=-10-5y+4z now we want to substitute this into the other two equations we have equation 2 and 3 2x-y+5z=-9 equation 2 2x-10y-5z=0 equation 3 these become 2(-10-5y+4z)-y+5z=-9 equation 2 2(-10-5y+4z)-10y-5z=0 equation 3 simplify these two get simultaneous equations with two unknowns -20-10y+8z-y+5z=-9 -11y+13z=11 equation 2 -20-10y+8z-10y-5z=0 -20y+3z=20 equation 3 put these side by side -11y+13z=11 equation 2 -20y+3z=20 equation 3 now rearrange one of these to get y the subject or z the subject -11y=11-13z y=(11-13z)/-11 substitute this into equation 3 -20(11-13z)/-11 + 3z=20 simplify (-220+260z)/-11 +3z=20 multiply by -11 to get rid of the fraction -220+260z -33z = -220 157z=0 z=0 now substitute z=0 into -11y+13z=11 equation 2 -11y+13(0)=11 -11y=11 y=-1 substitute z=0 and y=-1 into original equation x+5y-4z=-10 equation 1 x+5(-1)-4(0)=-10 x-5=-10 x=-5 just to make sure substitute x=-5 y=-1 z=0 into the other two original equations 2x-y+5z=-9 equation 2 2(-5)-(-1)+5(0)=-9 -10+1=-9 -9=-9 2x-10y-5z=0 2(-5)-10(-1)-5(0)=0 -10+10=0 0=0 x=-5 y=-1 z=0
Thank you so much!!! Would you be able to help me with one more?
does this help/ does this make sense?
Solve the system by substitution. 2x-y+z=-4 z=5 -2x+3y-z=-10 a.) (-8, 7, 5) b.) (–8, –7, 5) c.) (8, –7, 5) d.)(–8, –7, –5)
Can you help me with this one too? /:
you can do this... you are given the value of Z substitute it into the other 2 equations, then you will have two equations with two unknowns then you will rearrange one of the equations to make y the subject or x the subject lets say you make y the subject substitute y into the other equation you have then you will get a value for x then substitute x into one of the two equations you have to get y
Okk, let me try it. Then will you check my answers?
Ok.... I'm confused..
confused with what