I am 8th grade and I am confused by this question3x-4y=-15                5x+y=-2I really need help....

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You need to find a value of x and y that make both equations work. 3x - 4y = -15 5x +  y =   -2 is the same as saying: 3x - 4y = -15 20x + 4y = -8 (multiply every term in the equation by 4) 23x = -23      (add the equations together to eliminate the y term, so you can solve for x) x = -1 You can plug the x = -1 into the second equation to solve for y:  -5 + y = -2 or y = 3. Try it in the 1st equation -3 - 12 = -15 So your solution is x = -1 and y = 3

3x - 4y = -155x + y = -2First of all, notice that you have, in the second equation, y all by its lonesome, therefore, simply restate:y = - 5x -2, and then substitute into the top equation:3x - 4(-5x - 2) = -15; distribute -4 across the parentheses: 3x + 20x + 8 = -15; combine like elements: 23 x = -23, solve for x via multiplying inverse of 23: 23x(1/23) = -23(1/23); simplify: x = -1. Now plug the value of x into equation 3x - 4y = -15, (3)(-1) - 4y = -15; -3 -4y = -15; -4y = -12; y = 3. The solution set is: (-1, 3). For proof simply plug the values of x and y into both equations to make sure it produces true statements in both. Which it does. Another good approach would be to restate the equations into their slope-intercept forms; create a table of values for each; such as x = -3, -2, -1, 0, 1, 2, 3 for both equations and solve for y and then locate the shared set of values.