Add the equations: 4a + 4a - 5b + 5b = 7 + 17
ie 8a = 24
a = 3, so b = 1
8.00 − 5.91 = 2.09 (Method is to first subtract 2.09 from 8.00, which equals 5.91 If you then subtract 5.91 from 8.00 the answer will be 2.09)
x - y = 2 : Equation 1 2x + 3y = 4 : Equation 2 Multiplying Equation 1 by 3 gives 3x - 3y = 6 : Equation 3 Adding Equation 2 to Equation 3 gives 5x = 10 Dividing both sides by 5 gives x = 5 Substituting x=5 into Equation 1 gives 5 - y = 2 Therefore y = 3. Our final answer is therefore x=5 and y=3
Substitution method: from first equation y = 5x - 8. In the second equation this gives 25x - 5(5x - 8) = 32 ie 25x - 25x + 40 = 32 ie 40 = 32 which is not possible, so the system has no solution. Multiplication method: first equation times 5 gives 25x - 5y = 40, but second equation gives 32 as the value of the identical expression. No solution.
5x + 3y = -63x - 2y = 4 (multiply by 2 the first equation, multiply by 3 the second equation)10x + 6y = -129x - 6y = 12 (add both the equations)19x = 0 (divide by 19 both sides)x = 0 (substitute 0 for x into the first equation)5x + 3y = -65(0) + 3y = -63y = -6 (divide by 3 to both sides)y = -2Thus, (0, -2) is the solution of the given system of the equations.
I'll assume the simplified case of two equations, with two variables each. Some of the methods can be extended to more complicated cases.Substitution: Solve for one variable in one equation, replace it in the other equation.Setting two quantities equal: For example, if 5x + 3y = 10, and 5x - 2y = 0, solve each equation for "5x", and set the two equal, with the result: 10 - 3y = 2y.Addition/subtraction: Add or subtract one equation (or a multiple of one equation) to the other. In the previous example, if you subtract the second equation from the first, you get an equation that doesn't contain x.In any of these cases, after solving for a single variable, replace in one of the original equations to get the other variable.
x=1, y=1
Double first equation: 2x + 2y = 4 Subtract this from second equation giving 5y = 5 so y = 1 and x = 1
You need two equations to use the addition method.
solve system equation using addition method 3x-y=9 2x+y=6
From first equation, -y = 3x + 3. Substitute in second equation: -3x + 5(3x + 3) = -21 ie 12x = -36 so x = -3 and y = -(-9 + 3) = 6. Easier method: subtract first equation from second giving -4y = -24 so y = 6, this in first equation gives -6 = 3x + 3, ie 3x = -9 so x = -3
8.00 − 5.91 = 2.09 (Method is to first subtract 2.09 from 8.00, which equals 5.91 If you then subtract 5.91 from 8.00 the answer will be 2.09)
x+y=5
the answer
z=pq
The answer depends on the equation: there is no single method which can be used for all equations.
x - y = 2 : Equation 1 2x + 3y = 4 : Equation 2 Multiplying Equation 1 by 3 gives 3x - 3y = 6 : Equation 3 Adding Equation 2 to Equation 3 gives 5x = 10 Dividing both sides by 5 gives x = 5 Substituting x=5 into Equation 1 gives 5 - y = 2 Therefore y = 3. Our final answer is therefore x=5 and y=3
how do you use the substitution method for this problem 2x-3y=-2 4x+y=24