Question

# Solve for a and b $$\displaystyle 2\left ( a+b \right )-\left ( a-b \right )=6,\: 4\left ( a-b \right )=2\left ( a+b \right )-9$$

A
a=34,b=134
B
a=12,b=112
C
a=134,b=34
D
a=32,b=34

Solution

## The correct option is C $$\displaystyle a=\frac{3}{4},b=1\frac{3}{4}$$Given,$$2\left( a+b \right) -\left( a-b \right) =6$$$$4\left( a-b \right) =2\left( a+b \right) -9$$$$2\left( a+b \right) -\left( a-b \right) =6$$$$2a+2b-a+b=6$$$$a+3b=6$$ ........ $$(1)$$$$4\left( a-b \right) =2\left( a+b \right) -9$$$$4a-4b=2a+2b-9$$$$2a-6b=-9$$ ...... $$(2)$$On adding equations $$(1)$$ and $$(2)$$, we get$$2a+6b=12$$ $$2a-6b=-9$$__________________$$4a=3$$ $$a=\dfrac { 3 }{ 4 }$$Putting the value of $$a$$ in eq $$(1)$$, we get$$\dfrac { 3 }{ 4 } +3b=6$$$$3b=6-\dfrac { 3 }{ 4 }$$$$b=\dfrac { 21 }{ 4 } \times \dfrac { 1 }{ 3 } =\dfrac { 21 }{ 12 } =1\dfrac { 9 }{ 12 } =1\dfrac { 3 }{ 4 }$$Hence, $$a=\dfrac { 3 }{ 4 }$$ and $$b=1\dfrac { 3 }{ 4 }$$Maths

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