Question

A system of equations is given.

Equation 1: 5x − 2y = 10
Equation 2: 4x − 3y = 15

Explain how to eliminate x in the system of equations.

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Answer 1

To eliminate x, you need a positive coefficient in front of x for one equation and its negative counterpart in front of the other equation as a positive number plus its negative opposite equals 0 (e.g., -4 + 4 = 0 and -80 + 80 = 0)

Step 1: Therefore, we can eliminate x by first determining the least common multiple (LCM) between 5 and 4. We know that 5 * 4 = 20 and 4 * 5, so the LCM between 5 and 4 is 20.

Step 2: In order to have 20 as coefficient for x in one equation and -20 for x as a coefficient in the other equation, we can multiply the entire first equation by 4 and the entire second equation by -5:

Equation 1 multiplied by 4: 4 * (5x - 2y = 10) = 20x - 8y = 40

Equation 2 multiplied by -5: -5* (4x - 3y = 15) = -20x + 15y = -75

Step 3: Adding the two equations shows that the xs cancel as 20x - 20x = 0, leaving us with 15y - 8y = 40 - 75, which simplifies to 7y = -35

Answer 2

Answer: See below.

Explanation:

First, we are already given these equations in standard form.

5x − 2y = 10

4x − 3y = 15

Next, we need to make the coefficients of the x variables opposites (as in 5 and -5, etc), since we want to eliminate the x's. To do this, we will find a common multiple (here, the Lowest Common Multiplb is 20). Then, we will multiply every term by the number that makes the coefficient of x our common multiple.

We will make the first equation with a coefficient of 20 for the x and the second with a coefficient of -20 for the x.

See this visually below.

5x − 2y = 10 ➜ 4(5x) − 4(2y) = 4(10) ➜ 20x - 8y = 40

4x − 3y = 15 ➜ -5(4x) − -5(3y) = -5(15) ➜ -20x + 15y = -75

Lastly, add these two equations together. The x's are eliminated. This also will let us solve for y.

20x - 8y = 40

+ -20x + 15y = -75

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7y = -35

y = -5