Follow the directions to solve the system of equations by elimination. 8x + 7y = 39 4x – 14y = –68 Multiply the first equation to enable the elimination of the y-term. Add the equations to eliminate the y-terms. Solve the new equation for the x-value. Substitute the x-value back into either original equation to find the y-value. Check the solution. The solution to the system of equations is (, ).

8x +7y = 39 multiply everything by 2
2(8x +7y = 39)
16x +14y = 78
+4x - 14y= -68
-------------------------
20x = 10
20x /20 = 10/20
x = 1/2

Plug x into first formula
8 x + 7y =39
8(1/2) 7y=39
4+ 7y =39
4-4+7y =39-4
7y =35
7y/7 = 35/7
y = 5
(1/2, 5)

Answer Link

alokdubeyvidyaatech alokdubeyvidyaatech · Answer 2 · 2021-11-08T10:34:58-05:00

The given system of linear equations has a unique solution. The solution to the system of equations is \((\frac{1}{2}, 5)\).

Given equations:

\(8x+7y=39\) ...........(i)

\(4x-14y=-68\) ...........(ii)

Multiply the first equation by 2 to enable the elimination of the y-term.

\(2(8x+7y=39)\\16x+14y=78\) ...........(iii)

Add the equations (ii) and (iii) to eliminate the y-terms

\(4x-14y=-68\\16x+14y=78\)

------------------------------

\(20x=10\)

or \(x=\frac{10}{20}\)

or \(x=\frac{1}{2}\)

Plug x into equation (i) to find the x-value

\(8(\frac{1}{2}) +7y=39\\4+7y=39\\7y=39-4\\7y=35\\y=\frac{35}{7} \\y=5\)

Check the solution

\(8x+7y=39\)

For

\(x=\frac{1}{2} \\y=5\)

\(8(\frac{1}{2} )+7(5)=39\\4+35=39\\39=39\)

LHS = RHS

Therefore, the solution to the system of equations is \((\frac{1}{2} , 5)\).

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