les-grizzlys-catalans.org->Polynomials-and-rational-expressions-> SOLUTION: Here's the question and part of the solution.How many ordered pairs (x,y) satisfy this system of equations:x = 2y + 5y = (2x-3)(x+9)Here's part of the solution:Step 1 var visible_logon_form_ = false;Log in or register.Username: Password: Register in one easy step!.Reset your password if you forgot it."; return false; } "> Log On


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Click here to see ALL problems on Polynomials-and-rational-expressionsQuestion 978100: Here"s the question and part of the solution.How many ordered pairs (x,y) satisfy this system of equations:x = 2y + 5y = (2x-3)(x+9)Here"s part of the solution:Step 1. Substitute y in the 2nd equation into the first equation to get x = 2((2x-3)(x+9)) + 5Step 2. The above = 4x^2 + 30x - 54Step 3. The solution says this can be rewritten as: 4x^2 + 29x -54 = 0. Here are my problems: What happened to the 5 on the end of the equation in Step 1? Where did the 29 come from in Step 3? Thank you for any help?Helen Found 3 solutions through josgarithmetic, Boreal, MathTherapy:Answer by josgarithmetic(36412) (Show Source): You can put this solution on YOUR website! , the substitution for x.Find the solutions for y, which might be none, or one, or two of them.I CHOSE to make the substitution for x instead of for y. You could just as well make the sub for y as you may have wanted. My view is that the subst for x is easier to use.What you actually tried to do is tougher, and may be more work. Answer by Boreal(14371) (Show Source): You can put this solution on YOUR website! (2x-3)(x+9)= 2x^2+15x-272 times that is 4x^2+30x-54but x=4x^2-30x-54> ;; original equation was x=2y+5subtract x from both sides0=4x^2+29x-49You are right about the 5. That has to be carried through. Answer by MathTherapy(9881) (Show Source): You can put this solution on YOUR website! Here"s the question and part of the solution.How many ordered pairs (x,y) satisfy this system of equations:x = 2y + 5y = (2x-3)(x+9)Here"s part of the solution:Step 1. Substitute y in the 2nd equation into the first equation to get x = 2((2x-3)(x+9)) + 5Step 2. The above = 4x^2 + 30x - 54Step 3.

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The solution says this can be rewritten as: 4x^2 + 29x -54 = 0. Here are my problems: What happened to the 5 on the end of the equation in Step 1? Where did the 29 come from in Step 3? Thank you for any help?Helenx = 2y + 5 ----- eq (i)y = (2x - 3)(x + 9) ----- eq (ii)x = 2<(2x - 3)(x + 9)> + 5 ------- Substituting (2x - 3)(x + 9) for y in eq (i)
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