Question

Solve the given system of equations utilizing either the substitution or addition method.

Answer 1

Answer:

Value of x=4 and y=-13/3

Step-by-step explanation:

We need to solve the systems of equations

$5x+3y=7 \ and \ \frac{3}{2}x-\frac{3}{4}y =9\frac{1}{4}$

Solving using substitution method:

Let:

$5x+3y=7---eq(1) \\ \frac{3}{2}x-\frac{3}{4}y =9\frac{1}{4}---eq(2)$

Find value of x from eq(1) and put in eq(2)

$5x+3y=7\\5x=7-3y\\x=\frac{7-3y}{5}$

Put value of x in eq(2)

$\frac{3}{2}(\frac{7-3y}{5}) -\frac{3}{4}y =\frac{37}{4}\\\frac{21-9y}{10} -\frac{3}{4}y =\frac{37}{4}\\Taking \ LCM\\\frac{21*2-9y*2-3y*5}{20} =\frac{37}{4}\\\frac{42-18y-15y}{20} =\frac{37}{4}\\42-33y=\frac{37}{4}*20\\42-33y=185\\-33y=185-42\\-33y=143\\y=\frac{143}{-33}\\y=-\frac{13}{3}$

Now, finding value of x by putting value of y in eq(1)

$x=\frac{7-3y}{5}\\x= \frac{7-3(-\frac{13}{3}) }{5}\\x=\frac{7+13}{5}\\x=\frac{20}{5}\\x=4$

So, Value of x=4 and y=-13/3