-3x = 3 + y
3 + y - 5y = -21
-4y = -24
y = 6
x = -3
That's exactly the purpose of the substitution method ... to get an equation with one less variable. When you have it, you solve it for the variable that's left.
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.
y= -4x-7 y=3x 3x= -4x-7 3x+4x=7 7x=7 x=1
From Equation 1: x = 3y - 23Substitute for x in equation 2: 5(3y - 23) + 6y = 74ie 15y - 115 + 6y = 74ie 21y - 115 = 74Add 115 to each side: 21y = 189Divide each side by 21: y = 9x = 3y - 23 ie 27 - 23 ie 4x = 4, y = 9
3x+y = 10 y = x-2 Substitute the value of y into the top equation: 3x+x-2 = 10 => 4x = 10+2 => 4x = 12 => x = 3 Substitute the value of x into the original equations to find the value of y: So: x = 3 and y = 1
by elimination,substitution or through the matrix method.
x+y=5
how do you use the substitution method for this problem 2x-3y=-2 4x+y=24
-2
To solve this system of equations, we can use the method of substitution or elimination. Let's use the substitution method. From the second equation, we can express y as y = 55 - 4x. Substitute this expression for y in the first equation: 7x - 5(55 - 4x) = 76. Simplify this equation to solve for x. Then, substitute the value of x back into one of the original equations to find the value of y.
That's exactly the purpose of the substitution method ... to get an equation with one less variable. When you have it, you solve it for the variable that's left.
2x-3y=13
This is not Calculus.y=7(Already solved)substiute y=7 into y=8xtherefore 7 = 8xtherefore x = 7/8
There are no disadvantages. There are three main ways to solve linear equations which are: substitution, graphing, and elimination. The method that is most appropriate can be found by looking at the equation.
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.
From first equation: y = 2x - 5Substitute this in second equation: 3(2x - 5) - x = 5, ie 6x - 15 - x = 5ie 5x = 5 + 15 so x = 4 and y = 3
The substitution method in mathematics is a technique used to solve systems of equations by isolating one variable and substituting it into the other equation. The method is not attributed to a single inventor, as it has been used by mathematicians for centuries. The concept of substitution in algebra can be traced back to ancient civilizations such as the Babylonians and Greeks, who used similar methods to solve mathematical problems.