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What is the solution to y-2x 5 and x-2y-15?
Assuming y=2x+5 and x=2y-15y=2x+5 (1)x=2y-15 (2)Substitute y=2x+5 into (2):x=2(2x+5)-15x=4x+10-150=3x+10-150=3x-55=3x5/3=x (x=5/3)Substitute x=5/3 into y=2x+5:y=2(5/3)+5y=10/3+5y=25/3x=5/3, y=25/3
Factor 6x2 plus 11x-10?
6x2 + 11x - 10 = (6x2 + 15x) - (4x + 10) = 3x(2x + 5) - 2(2x + 5) = (3x - 2)(2x + 5). Note, first, that the three co-efficients of the original quadratic expression are 6, 11, and -10. We need two numbers whose sum is 11 and whose product is (6)(-10) = -60. These two numbers turn out to be 15 and -4. Thus, we replace the middle term, 11x with 15x - 4x and proceed with the factorisation in the usual way.
What is the answer to -2x y5?
-2x = y^5 or y^5 = -2x y = (-2x)^(1/5)
How do you change 15x-y-45 into slope intercept?
5y = 15x -45 Divide by 5 y = 15x/5 -45/5 y = 3x -9 y = mx + b m= 3 6 = -9 The slope is 3 The y intercept is -9
Solve by substitution method 15x-20y7 5y-3 10x?
15x - 20y = 7 5y - 3 = 10x Let's solve the second equation for y. 5y - 3 = 10x add 3 to both sides; 5y = 10x + 3 divide by 5 to both sides; y = 2x + 3/5 Substitute the expression 2x + 3/5 for y to the first equation. 15x - 20y = 7 15x - 20(2x + 3/5) = 7 multiply by 20 each term of the parentheses; 15x -40x - 60/5 = 7 - 25x - 12 = 7 add 12 to both sides; -25x = 19 multiply by -1 to both sides; 25x = -19 divide by 25 to both sides; x = -19/25 Substitute -19/25 for x to y = 2x + 3/5 y = 2x + 3/5 y = 2(-19/25) + 3/5 y = -38/25 + 3/5 multiply 3/5 by 5/5; y = -38/25 + 15/25 y =-23/25 The solution set is (-19/25, -23/25) Check: 1. 15 x - 20y = 7 15(-19/25) - 20(-23/25) = 7? -285/25 + 460/25 = 7 ? 175/25 = 7 ? 7 = 7 2. 5y - 3 = 10x 5(-23/25) - 3 = 10(-19/25) ? -115/25 - 3 = -190/25 ? -115/25 - 3(25/25) = -190/25 ? -115/25 - 75/25 = -190/25 ? -190/25 = -190 / 25 or, -38/5 = -38/5 The solution set is (1/5, 1)
