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Multiply the first equation by 3 to give 9x + 15y = 21

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Q: How can you get coefficients that are equal for y. 3x plus 5y equals 7 2x plus 15y equals 11?
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What are the solutions to the simultaneous equations of 2x -3y equals 4 and 5x plus 2y equals 1?

Equations: 2x-3y = 4 and 5x+2y = 1 Multiply all terms in 1st equation by 5 and all terms in 2nd equation by 2:- So: 10x-15y = 20 and 10x+4y = 2 Subtract the 2nd equation from the 1st equation: -19y = 18 => y = -18/19 By substitution the solutions are: x = 11/19 and y = -18/19


3x plus 6y equals 18 2x plus 5y equals 20?

You need to be more clear about your question. 3x + 6y = 18 2x + 5y = 20 The two equations you give represent lines. I assume you're trying to find the point at which they intersect. There are various ways to work this out. One technique is to multiply each equation by a factor that gives one of the two variables the same coefficient. For example, if we multiply the first equation by 2, and the second equation by 3, we get: 6x + 12y = 36 6x + 15y = 60 Then we can subtract one from the other, which gives us a solution for y: [6x + 12y = 36] - [6x + 15y = 60] _______________ [0x - 3y = -24] So -3y = -24, telling us that at the point of intersection, y = 8. We can then plug that back into one of the original equations and solve for x: 2x + 5y = 20 2x + 5(8) = 20 2x = 20 - 40 2x = -20 x = -10 So the lines intersect at the point (-10, 8). Another way to work it out is to solve one of the equations for a single variable, and then plug it into the other one: 3x + 6y = 18 ∴ x + 2y = 6 ∴ x = 6 - 2y 2x + 5y = 20 ∴ 2(6 - 2y) + 5y = 20 ∴ 12 - 4y + 5y = 20 ∴ y = 20 - 12 ∴ y = 8 That gives us the value 8 for y, which we can once again plug in to either of the original equations to find x: 2x + 5y = 20 ∴ 2x + 40 = 20 ∴ 2x = -20 ∴ x = -10 Once again telling us that the lines intersect at the point (-10, 8)