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Question: How do you solve the system by graphing the linear system \(2x-y=5\) and \(3x+2y=4\) ?...

How do you solve the system by graphing the linear system 2xy=52x-y=5 and 3x+2y=43x+2y=4 ?

Explanation

Solution

The above 2 equations are of straight lines. The solution to the system of equations will be the intersection between 2 lines. There will be no solution if 2 lines are parallel to each other and infinite solution if both lines are the same.

Complete step-by-step answer:
General formula for the line is y=mx+cy=mx+c
2xy=52x-y=5 and 3x+2y=43x+2y=4 are equations of straight lines. The intersection point will be the solution of the system.
General equation for the line is y=mx+cy=mx+c where m is the slope and c is the y intercept, we can write 2xy=52x-y=5 as y=2x5y=2x-5 and if we compare y=2x5y=2x-5 with y=mx+cy=mx+c we get the slope is 2 and y intercept is equal to – 5, in 3x+2y=43x+2y=4 the slope is 32-\dfrac{3}{2} and y intercept is equal to 2 so let’s draw the graph and find out the point of intersection.


We can see that the point of intersection is A(2 , - 1) x coordinate is the solution. So 2 is the solution.

Note: Another method is to solve by determinant method
Suppose there are 2 linear equation a1x+b1y=c1{{a}_{1}}x+{{b}_{1}}y={{c}_{1}} and a2x+b2y=c2{{a}_{2}}x+{{b}_{2}}y={{c}_{2}} the solution to this problem let’s take
D= a1b1 a2b2 \left| \begin{matrix} {{a}_{1}} & {{b}_{1}} \\\ {{a}_{2}} & {{b}_{2}} \\\ \end{matrix} \right|
A= b1c1 b2c2 \left| \begin{matrix} {{b}_{1}} & {{c}_{1}} \\\ {{b}_{2}} & {{c}_{2}} \\\ \end{matrix} \right|
B= a1c1 a2c2 \left| \begin{matrix} {{a}_{1}} & {{c}_{1}} \\\ {{a}_{2}} & {{c}_{2}} \\\ \end{matrix} \right|
x=ADx=-\dfrac{A}{D} and y=BDy=\dfrac{B}{D} Where D0D\ne 0
If D=0D=0 and any one of A and B is not equal to 0 then there will be no solution for the system of equation
If D=A=B=0D=A=B=0 there will be infinitely many solutions to the system of equations.
If D0D\ne 0 there will be one solution for the system of equations.
In this case a1=2,b1=1,c1=5{{a}_{1}}=2,{{b}_{1}}=-1,{{c}_{1}}=5 and a2=3,b2=2,c2=4{{a}_{2}}=3,{{b}_{2}}=2,{{c}_{2}}=4
Solving the determinant D = 7 , A= - 14 and B = -7
So x=ADx=-\dfrac{A}{D}
So x = 2

And y=BDy=\dfrac{B}{D}= - 1
While solving the system of linear equations we can imagine the equation having 2 unknown as equations of straight in 2-D Cartesian plane so 2 equations means 2 lines will intersect at 1 point unless both lines are parallel or both lines are overlapping. Same goes with 3 unknowns: we can imagine the equations as equations of plane in 3D geometry and then solve for the unknown variables.