Question

Systems of equations and inequalities

solving systems by elimination
{y= -x+1
{y= 4x-14

Answer 1

x = 3 , y = -2

Explanation:

Solve the following system:

{y = 1 - x | (equation 1)

y = 4 x - 14 | (equation 2)

Express the system in standard form:

{x + y = 1 | (equation 1)

-(4 x) + y = -14 | (equation 2)

Swap equation 1 with equation 2:

{-(4 x) + y = -14 | (equation 1)

x + y = 1 | (equation 2)

Add 1/4 × (equation 1) to equation 2:

{-(4 x) + y = -14 | (equation 1)

0 x+(5 y)/4 = (-5)/2 | (equation 2)

Multiply equation 2 by 4/5:

{-(4 x) + y = -14 | (equation 1)

0 x+y = -2 | (equation 2)

Subtract equation 2 from equation 1:

{-(4 x)+0 y = -12 | (equation 1)

0 x+y = -2 | (equation 2)

Divide equation 1 by -4:

{x+0 y = 3 | (equation 1)

0 x+y = -2 | (equation 2)

Collect results:

Answer: {x = 3 , y = -2

Answer 2

x = 3, y = -2 ⇒ (3, -2)

Explanation:

`\left\{\begin{array}{ccc}y=-x+1\\y=4x-14&\text{chnge the signs}\end{array}\right\\\\\underline{+\left\{\begin{array}{ccc}y=-x+1\\-y=-4x+14\end{array}\right}\qquad\text{add both sides of the equations}\\.\qquad0=-5x+15\qquad\text{add}\ 5x\ \text{to both sides}\\.\qquad5x=15\qquad\text{divide both sides by 5}\\.\qquad(5x)/(5)=(15)/(5)\\.\qquad\boxed{x=3}\\\\\text{Substitute the value of}\ x\ \text{to the first equation:}\\\\y=-3+1\\\boxed{y=-2}`