To solve a pair of equations using substitution, first solve one of the equations for one of the variables. Then substitute the result for that variable in the other equation.

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Choose one of the equations and solve it for x by isolating x on the left hand side of the equal sign.

Substitute -frac73 for y in x=3y+5. Because the resulting equation contains only one variable, you can solve for x directly.

6x-3y=15;x+3y=-8 Solution : x,y = 1,-3 System of Linear Equations entered : <1> 6x - 3y = 15 <2> x + 3y = -8 Graphic Representation of the Equations : -3y + 6x = 15 3y + x = -8 Solve by ...

6x-4y=39;2x+y=6 Solution : x,y = 9/2,-3 System of Linear Equations entered : <1> 6x - 4y = 39 <2> 2x + y = 6 Graphic Representation of the Equations : -4y + 6x = 39 y + 2x = 6 Solve by ...

displaystylex=-frac1811 ,displaystyley=-frac6455 Explanation:To solve by elimination you will need to make coefficients of x or y "match" so that when you add or ...

6x-3y=5;2x+y=2 Solution : x,y = 11/12,1/6 System of Linear Equations entered : <1> 6x - 3y = 5 <2> 2x + y = 2 Graphic Representation of the Equations : -3y + 6x = 5 y + 2x = 2 Solve through ...

2x-3y=5;x+2y=-1 Solution : x,y = 1,-1 System of Linear Equations entered : <1> 2x - 3y = 5 <2> x + 2y = -1 Graphic Representation of the Equations : -3y + 2x = 5 2y + x = -1 Solve by ...

By substitution.Explanation:We have two equations,displaystyle5x+5y=27anddisplaystyle9x-5y=0 . In order to use substitution, we need to solve for one variable. Tthe ...

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To solve a pair of equations using substitution, first solve one of the equations for one of the variables. Then substitute the result for that variable in the other equation.

Choose one of the equations and solve it for x by isolating x on the left hand side of the equal sign.

Substitute -frac73 for y in x=3y+5. Because the resulting equation contains only one variable, you can solve for x directly.

left(eginmatrix1&-3\5&-3endmatrix
ight)left(eginmatrixx\yendmatrix
ight)=left(eginmatrix5\-3endmatrix
ight)

inverse(left(eginmatrix1&-3\5&-3endmatrix
ight))left(eginmatrix1&-3\5&-3endmatrix
ight)left(eginmatrixx\yendmatrix
ight)=inverse(left(eginmatrix1&-3\5&-3endmatrix
ight))left(eginmatrix5\-3endmatrix
ight)

Left multiply the equation by the inverse matrix of left(eginmatrix1&-3\5&-3endmatrix
ight).

left(eginmatrix1&0\0&1endmatrix
ight)left(eginmatrixx\yendmatrix
ight)=inverse(left(eginmatrix1&-3\5&-3endmatrix
ight))left(eginmatrix5\-3endmatrix
ight)

left(eginmatrixx\yendmatrix
ight)=inverse(left(eginmatrix1&-3\5&-3endmatrix
ight))left(eginmatrix5\-3endmatrix
ight)

left(eginmatrixx\yendmatrix
ight)=left(eginmatrixfrac-3-3-left(-3 imes 5
ight)&-frac-3-3-left(-3 imes 5
ight)\-frac5-3-left(-3 imes 5
ight)&frac1-3-left(-3 imes 5
ight)endmatrix
ight)left(eginmatrix5\-3endmatrix
ight)

For the 2 imes 2 matrix left(eginmatrixa&b\c&dendmatrix
ight), the inverse matrix is left(eginmatrixfracdad-bc&frac-bad-bc\frac-cad-bc&fracaad-bcendmatrix
ight), so the matrix equation can be rewritten as a matrix multiplication problem.

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left(eginmatrixx\yendmatrix
ight)=left(eginmatrix-frac14¼\-frac512&frac112endmatrix
ight)left(eginmatrix5\-3endmatrix
ight)

left(eginmatrixx\yendmatrix
ight)=left(eginmatrix-frac14 imes 5+frac14left(-3
ight)\-frac512 imes 5+frac112left(-3
ight)endmatrix
ight)

In order to solve by elimination, coefficients of one of the variables must be the same in both equations so that the variable will cancel out when one equation is subtracted from the other.