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 ...
More Items     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.      