Question

#1: Solve the linear system below using the elimination method. Type your

Answer 1

Answer:

$(-3,-2)$

Step-by-step explanation:

First, you need to get one like term in each equation the same in order to eliminate. To do less damage to the equation, we'll use the x terms.

To get the x terms the same, multiply both equations by the opposite x term:

$-3(2x+5y)=-3(-16)\\\\2(-3x+7y)=2(-5)$

Simplify by multiplying. Use the distributive property for the left side (remember, two negatives make a positive and a negative multiplied by a positive will always be negative):

$-3(2x)-3(5y)=-3(-16)\\\\2(-3x)+2(7y)=2(-15)\\\\\\-6x-15y=48\\\\-6x+14y=-10$

Now subtract all like terms. The x terms cancel out:

$-6x-(-6x)=0\\\\-15y-14y=-29y\\\\48-(-10)=58\\\\-29y=58$

Now solve the new equation for y. Isolate the variable by using inverse operations. The term can be seen as -29×y. In order to isolate the variable, use division (opposite of multiplication). Divide both sides by 29:

$\frac{-29y}{29} =\frac{58}{29} \\\\-1y=2$

Make the variable positive. Multiply both sides by -1:

$-1(-1y)=-1(2)\\\\y=-2$

The value of y is -2. Insert this value into either original equation:

$2x+5(-2)=-16$

Solve for x. Simplify multiplication:

$2x-10=-16$

Use inverse operations to get like terms on the same side of the equation. Add 10 to both sides (addition is the opposite of subtraction) to cancel out the -10 on the left:

$2x-10+10=-16+10\\\\2x=-6$

Isolate the variable using inverse operations. The x term can be seen as 2×x. Division is the opposite of multiplication, so divide both sides by 2:

$\frac{2x}{2}=\frac{-6}{2} \\\\x=-3$

The value of x is -3. Therefore, the solution to the system is:

$(-3_{x},-2_{y})$

:Done