3w – 4z = 8 2w + 3z = -6 solve
The correct answer and explanation is:
We are given a system of two linear equations:
- 3w−4z=83w – 4z = 8
- 2w+3z=−62w + 3z = -6
Step 1: Use the method of elimination to eliminate one variable.
We want to eliminate one variable by making the coefficients of either ww or zz the same (or opposites). Let’s eliminate zz.
To do this, find the least common multiple (LCM) of the coefficients of zz:
- LCM of 4 and 3 is 12.
We’ll multiply each equation to get zz terms to be opposites:
Multiply equation (1) by 3: 3(3w−4z)=3(8)⇒9w−12z=24(3)3(3w – 4z) = 3(8) \Rightarrow 9w – 12z = 24 \tag{3}
Multiply equation (2) by 4: 4(2w+3z)=4(−6)⇒8w+12z=−24(4)4(2w + 3z) = 4(-6) \Rightarrow 8w + 12z = -24 \tag{4}
Step 2: Add equations (3) and (4) to eliminate zz:
(9w−12z)+(8w+12z)=24+(−24)(9w – 12z) + (8w + 12z) = 24 + (-24)
Simplify: 17w=0⇒w=017w = 0 \Rightarrow w = 0
Step 3: Substitute w=0w = 0 into one of the original equations to find zz:
Use equation (1): 3w−4z=8⇒3(0)−4z=8⇒−4z=8⇒z=−23w – 4z = 8 \Rightarrow 3(0) – 4z = 8 \Rightarrow -4z = 8 \Rightarrow z = -2
✅ Final Answer:
w=0,z=−2\boxed{w = 0, \quad z = -2}
✍️ 300-Word Explanation:
This problem involves solving a system of linear equations with two variables using the elimination method, a strategy to remove one variable so we can solve for the other.
We started with:
- 3w−4z=83w – 4z = 8
- 2w+3z=−62w + 3z = -6
To eliminate one variable, we decided to eliminate zz. The goal was to manipulate the equations so the coefficients of zz become opposites. The least common multiple (LCM) of 4 and 3 is 12, so we multiplied the equations to get −12z-12z and +12z+12z. Specifically:
- Multiply the first equation by 3 to get 9w−12z=249w – 12z = 24
- Multiply the second equation by 4 to get 8w+12z=−248w + 12z = -24
Adding these two new equations cancels the zz terms, leaving a single-variable equation: 9w+8w=17w=0⇒w=09w + 8w = 17w = 0 \Rightarrow w = 0
Substituting w=0w = 0 back into the original equation allows us to solve for zz: 3(0)−4z=8⇒z=−23(0) – 4z = 8 \Rightarrow z = -2
This solution means the values w=0w = 0 and z=−2z = -2 satisfy both original equations, making it the unique solution to the system. Solving systems like this is important in algebra and real-world applications such as economics, physics, and engineering, where multiple conditions must be satisfied simultaneously.