Question

Solve the systems of equations:

$\frac{x-2}{4}+\frac{y+1}{3}=2;\frac{x+1}{7}+\frac{y-3}{2}=\frac{1}{2}$

• A. $(6,2)$
• B. $(2,2)$
• C. $(2,3)$
• D. $(3,4)$

Answer 1

Answer: (A)

$(6,2)$

$\frac{x-2}{4}+\frac{y+1}{3}=2$ ......... (1)

$\frac{x+1}{7}+\frac{y-3}{2}=\frac{1}{2}$ ..... (2)

First, we need to simplify the two equations. Let's multiply both sides of equation (1) by 12 and simplify.

$12(\frac{x-2}{4}+\frac{y+1}{3})=12\left(2\right)$

$3(x-2)+4(y+1)=24$

$3x-6+4y+4=24$

$3x+4y-2=24$

$3x+4y=26$

Let's multiply both sides of eq. (2) by 14

$14(\frac{x+1}{7}+\frac{y-3}{2})=14\left(\frac{1}{2}\right)$

$2(x+1)+7(y-3)=7$

$2x+2+7y-21=7$

$2x+7y-19=7$

$2x+7y=26$

Now we have the following system to solve

$3x+4y=26$ .............. (3)

$2x+7y=26$ ............... (4)

Because changing the form of their eq (3) or eq. (4) in preparation for the substitution method would produce a fractional form, let's use the elimination-by-addition method. We can start by multiplying equation (4) by -3

$3x+4y=26$ ............. (5)

$-6x-21y=-78$ .......... (6)

No, we can replace eq. (6) with an eq. we form by multiplying eq. (5) by 2 then adding that result to eq. (6)

$3x+4y=26$ ............ (7)

$-13y=-26$ ............ (8)

From eq. (8) we can find the value of y.

$-13y=-26\Rightarrow y=2$

Now we can substitute 2 for y in eq. (7)

$3x+4y=26$

$3x+4\left(2\right)=26$

$3x=18\Rightarrow x=6$

The solution set is $(6,2).$