i only need help on these two:
*2/3y=10+4x
5x=1/3y-8
*5x=4y-30
2x+3y=-12
DIRECTIONS:use linear combinations to solve the system of linear equations.. thanks :)
I suggest you study the two answers given before and proceed with solving the remainder. This way, you will have no problems to face your exams.
One of the answers is:
http://www.jiskha.com/display.cgi?id=1295847952
Have you tried substitution?
In the second problem,
x = -6 - (3/2)y
-30 -(15/2)y = 4y -30
y = 0
5x = -30
x = -6
To solve the system of linear equations using linear combinations, you need to follow these steps:
Step 1: Choose one equation and multiply both sides by a suitable number so that the coefficients of one variable will cancel out when combined with the other equation.
Step 2: Choose the other equation and multiply both sides by a different suitable number so that the coefficients of the same variable will cancel out when combined with the first equation.
Step 3: Add or subtract the two modified equations to eliminate one variable and solve for the other.
Step 4: Substitute the value found into one of the original equations to solve for the remaining variable.
Let's solve each of the systems of linear equations you provided using linear combinations:
System 1:
Equation 1: 2/3y = 10 + 4x (Equation A)
Equation 2: 5x = 1/3y - 8 (Equation B)
Step 1: Multiply Equation A by 3 to get rid of the fraction:
(3) * (2/3y) = (3) * (10 + 4x)
2y = 30 + 12x
Step 2: Multiply Equation B by 2 to get the coefficients of the y term to match:
(2) * (5x) = (2) * (1/3y - 8)
10x = 2/3y - 16
Step 3: Combine the modified equations:
2y = 30 + 12x
10x = 2/3y - 16
Multiply both sides of the second equation by 3 to get rid of the fraction:
30x = 2y - 48
Step 4: Now we have two equations in terms of y and x:
2y = 30 + 12x (Equation C)
30x = 2y - 48 (Equation D)
Subtract Equation D from Equation C:
2y - 2y = 30 + 12x - (2y - 48)
0 = 30 + 12x - 2y + 48
Combine like terms:
0 = 78 + 12x - 2y
Now rearrange the equation:
12x - 2y = -78 (Equation E)
Now we have a single equation involving x and y. You can solve Equation E using additional methods, such as substitution or elimination.
Repeat the steps above for System 2:
System 2:
Equation 1: 5x = 4y - 30 (Equation F)
Equation 2: 2x + 3y = -12 (Equation G)
Step 1: Multiply Equation F by 2:
(2) * (5x) = (2) * (4y - 30)
10x = 8y - 60
Step 2: No need to multiply Equation G since the coefficients already match.
Step 3: Combine the modified equations:
10x = 8y - 60
2x + 3y = -12
Multiply Equation F by 3 to match the coefficients of the y term:
30x = 24y - 180
Step 4: Now we have two equations in terms of y and x:
10x = 8y - 60 (Equation H)
30x = 24y - 180 (Equation I)
Multiply Equation H by 3:
30x = 24y - 180
Subtract Equation I from Equation H:
10x - 30x = 8y - 24y - 180
-20x = -16y - 180
Now rearrange the equation:
20x - 16y = -180 (Equation J)
Like before, Equation J can now be solved using additional methods, such as substitution or elimination.