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Algebra
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how can i solve a system of equations using substitution x+4y=-15 3x+3y-0
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for substitution look at the 2nd equation and rewrite it 3x + 3y = 0 or 3x = -3y or x = -y substitute this into the 1st equation (-y) + 4y = -15 or 3y = -15 solve for y when you have y substitute it into either equation to find x
so how do i solve 3y=-15 do i add 3y to both sides
divide both sides of the equation by 3
ok i got y=5 is that correct
not quite it should be y = -5
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so looking at the 1st equation and substituting y = -5 x + 4*(-5) = -15 so x - 20 = -15 whats x..?
oh k i got that one so how do i do 9x-8y= -6 x-3y=-7 do i do the same and how
so i subtract 20 and -15 right.
i am not sure
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