There is only one equation, so it's not a system of equations.
when x = 2
A possible first step in eliminating a variable in the system of equations (4x + 5y = -23) and (3x + 10y = -14) is to manipulate the equations to align the coefficients of one of the variables. For instance, you can multiply the first equation by 2 to obtain (8x + 10y = -46). This will allow you to eliminate the (y) variable by subtracting the second equation from the modified first equation.
it equals 28
-14
6x plus 3 equals 8x plus 14 when solved mathematically is equal to x = -5.5.
when x = 2
The equations are identical in value, ie the second is merely twice the first...
A possible first step in eliminating a variable in the system of equations (4x + 5y = -23) and (3x + 10y = -14) is to manipulate the equations to align the coefficients of one of the variables. For instance, you can multiply the first equation by 2 to obtain (8x + 10y = -46). This will allow you to eliminate the (y) variable by subtracting the second equation from the modified first equation.
why will the equations x+14=37 and x-14=37 have different solutions for x
It is not possible to solve one equation in two unknowns (x and y). Two independent equations are required.
it equals 28
4x - y = 1 3x + y = 13 Add the equations: 7x = 14 so that x = 2 and then, by the first equations, 4*2 - y= 1 so that y = 7
14 + 14 = 28
-14
6x plus 3 equals 8x plus 14 when solved mathematically is equal to x = -5.5.
13...... 9 plus 5 equals 14...... 14 minus 1 equals 13
x-2y=14 x+3y =9 subtract the equations x-2y = 14 -x-3y = -9 -5y = 5 divie both sides by -5 y = -1 substitute y = -1 into the first equation (or second) x -2(-1) = 14 x + 2 = 14 x = 12