By adding the original equations, eliminated x to find x = 6. Substituting into the first equation yielded y = -0.75. For the second system, solving by elimination gave p = -1 and h = 9. Verified solutions algebraically.
For the system {x + 4y = 3, 0.5x - 4y = 6}
1. Complete the statement:
- Adding the original equations in this system would result in eliminating the variable x.
2. Perform the selected operation:
- Adding the original equations:
(x + 4y) + (0.5x - 4y) = 3 + 6
Combine like terms:
1.5x = 9
Divide by 1.5:
x = 6
3. Solve the system using the equation from question 2:
Substitute x = 6 into the first equation:
6 + 4y = 3
Subtract 6 from both sides:
4y = -3
Divide by 4:
y = -0.75
4. Verify the solution algebraically:
Substitute x = 6 and y = -0.75 into both original equations:
- For the first equation: 6 + 4(-0.75) = 3, which is true.
- For the second equation: 0.5(6) - 4(-0.75) = 6, which is also true.
For the system {5p + 2h = 13, 3p - h = -12}
5. Solve the system using elimination:
Multiply the second equation by 2 to make the coefficients of h the same:
5p + 2h = 13
6p - 2h = -24
Now, add the first equation and the modified second equation to eliminate h :
11p = -11
Divide by 11:
p = -1
Substitute p = -1 into the second equation to solve for h:
3(-1) - h = -12
-3 - h = -12
Add 3 to both sides:
-h = -9
Multiply by -1:
h = 9
So, for the first system, the solution is
, and for the second system,
.