Solve the system of two equations using eliminations with multiplications

Question

4x - 6y = -48
-2x + 4y = 30

Answer 1

To solve the system of equations using elimination, we start with the two given equations:

1. $4x-6y=-48$ \quad (Equation 1)
2. $-2x+4y=30$ \quad (Equation 2)

Step 1: Multiply the equations to align the coefficients for elimination.

To eliminate $x$, we can manipulate the equations such that their coefficients are opposites. We can multiply Equation 2 by 2:

$$2\times (-2x+4y)=2\times 30$$

This gives us:

$$-4x+8y=60(Equation3)$$

Step 2: Now, we can add Equation 1 and Equation 3.

The updated equations are:

1. $4x-6y=-48$
2. $-4x+8y=60$

Now, we add both equations:

$$(4x-6y)+(-4x+8y)=-48+60$$

This simplifies to:

$$0x+2y=12$$

So, we have:

$$2y=12$$

Step 3: Solve for $y$.

Dividing both sides by 2, we get:

$$y=6$$

Step 4: Substitute $y$ back into one of the original equations to solve for $x$.

We can use Equation 1:

$$4x-6(6)=-48$$

This simplifies to:

$$4x-36=-48$$

Adding 36 to both sides gives:

$$4x=-48+36$$ $$4x=-12$$

Now we divide by 4:

$$x=-3$$

Step 5: Solution

The solution to the system of equations is:

$$\boxed{{\displaystyle (-3,6)}}$$

Verification

Let's verify by substituting $x=-3$ and $y=6$ back into the original equations:

1. Substituting into Equation 1: $$4(-3)-6(6)=-12-36=-48\text{(True)}$$

2. Substituting into Equation 2: $$-2(-3)+4(6)=6+24=30\text{(True)}$$

Both equations check out, confirming that the solution $(-3,6)$ is correct.