two systems of equations are shown. the first equation in system b is the original equation in system a. the…

two systems of equations are shown. the first equation in system b is the original equation in system a. the second equation in system b is the sum of that equation and a multiple of the second equation in system a. a. $\frac{1}{2}x + 3y = 11 \to \frac{1}{2}x + 3y = 11$, $5x - y = 17 \to 15x - 3y = 51$, $15\frac{1}{2}x = 62$. b. $\frac{1}{2}x + 3y = 11$, $15\frac{1}{2}x = 62$. what is the solution to both systems a and b? (3, 4) (3, 5) (4, 3) (5, 3)

two systems of equations are shown. the first equation in system b is the original equation in system a. the second equation in system b is the sum of that equation and a multiple of the second equation in system a. a. $\frac{1}{2}x + 3y = 11 \to \frac{1}{2}x + 3y = 11$, $5x - y = 17 \to 15x - 3y = 51$, $15\frac{1}{2}x = 62$. b. $\frac{1}{2}x + 3y = 11$, $15\frac{1}{2}x = 62$. what is the solution to both systems a and b? (3, 4) (3, 5) (4, 3) (5, 3)

Answer

Explanation:

Step1: Solve for x in the equation $15\frac{1}{2}x = 62$

Rewrite $15\frac{1}{2}$ as $\frac{31}{2}$. So the equation is $\frac{31}{2}x=62$. Multiply both sides by $\frac{2}{31}$: $x = 62\times\frac{2}{31}=4$.

Step2: Substitute x = 4 into the first - equation of system A or B

Substitute $x = 4$ into $\frac{1}{2}x+3y = 11$. We get $\frac{1}{2}\times4+3y = 11$. Simplify the left - hand side: $2 + 3y=11$. Subtract 2 from both sides: $3y=11 - 2=9$. Divide both sides by 3: $y = 3$.

Answer:

$(4,3)$