If you mean a slope of -5 and a point of (6, 3) then the equation is y = -5x+33
If the point is (6, 3) and the slope is -1/3 then the equation works out as 3y = -x+15
It is 13x + y = 81.
This is a set of three numbers. It is NOT an equation. There is nothing missing from a non-existent equation.
One equation will represent the number of coins, x + y = 63. The second equation will represent the value of the coins, 0.10x + 0.05y = 5.25. Solve the first equation for y (=63-x) and plug into the second equation. So, 0.1x + 0.05(63-x) = 5.25 Solve this and get x=42. your welcome mizz litta bay be aka anhellita richards
If you mean a slope of -5 and a point of (6, 3) then the equation is y = -5x+33
If you mean a slope of -12 through the point (5, 3) the equation is y = -12x+63
If the point is (6, 3) and the slope is -1/3 then the equation works out as 3y = -x+15
It is 13x + y = 81.
Points: (2, -5) and (6, 3) Slope: 2 Equation: y = 2x-9
Two coordinates of x and y are needed to work out the slope of a straight line equation.
63
When you graph the equation [ y = 7x ], the graph is a straight line that passes throughthe origin and has a slope of 7.The x-y coordinates of every point on the line are an ordered pair that satisfies the equation.There are as many of them as there are points on a line ... an infinite number.Here are a few of them:(0, 0)(1, 7)(0.2, 1.4)(-9, -63)(137, 959)
If you mean points of (5, 4) and (6, 3) then the slope is -1 and equation is y=-x+9
The required equation is: -7x = 63
7 times 9 is 63.
If you know the slope (m) and a point which we'll call x1, y1 then the equation of the line can be found by using the formula y - y1 = m(x - x1)Input the know values and then bring all the unknowns (x and y) to the left hand side and the knowns (the numbers) to the right hand side.For example:* Find the equation of the line that passes through the point (-2 , 5) and has a slope of -4. * Substitute y1 , x1 and m in the point slope form of a line * y - y1 = m(x - x1) * y - 5 = - 4(x - (-2))