Solve using substitution: -2x − y + 1 = 0 2x = -13 − 3y (1, -9) (3, -5) (-2, 5) (4, -7)
Solve for a variable. -2x − y + 1 = 0 -y = 2x - 1 y = -2x + 1 Substitute this back into the second equation as y. 2x = -13 − 3y 2x = -13 − 3(-2x + 1) 2x = -13 + 6x - 3 2x = -16 + 6x 4x = 16 x = 4 Plug x into the equation to solve for y. y = -2(4) + 1 y = -8 + 1 y = -7
i already solved this one for you, it's (4,-7)
Solve using substitution: y − 2x − 1 = 0 -4y = 6 − 8x (1/2, 2) (3, 1) No solution Infinitely many solutions
Let -2x-y=1. and 2x + 3y=-13. Thus, using substitution, let y =-2x-1..... becomes 2x + 3(-2x-1)=-13 therefore, x=4. y=-7.....
y = 2x +1 sub into -4y = 6 -8x -4(2x+1) = 6 -8x -8x -4 = 6 -8x the -8x's cancel -4 does NOT = 6 so no solution.
Select the choice below that could be the first step in solving the following system of equations using elimination: -4x − 10y = 3 3y − 3x − 7 = 0 Multiply the first equation by -4 and the second equation by 3 Multiply the first equation by 3 and the second equation by 10 Replace the right side of the first equation with the second equation Write the second equation in terms of x
multiply the first equation by 3 and the second by 10
Solve using elimination: 5x − 4y − 10 = 0 -3x − 4y = -6 (-2, -5) (2, 0) (-6, -6) (6, 5)
multiply the second equation by -1 and add 10 over to the other side in eq1 5x -4y =10 3x +4y = 6 (add) 8x = 16 x = 2 sub back in to get y 5(2) -4y = 10 10 -4y =10 (10's cancel) -4y = 0 y = 0 (2,0)
Solve using elimination: 7x − 3y − 1= 0 4y − 6x = 12 (6, 2) (7,18) (4, 9) (0, -1/3)
WHAT DO I DO
sorry, was gone for a bit.
first put all in standard form to make it easier 7x - 3y = 1 -6x + 4y = 12 next, multiply first by 4 and second by 3 28x - 12 = 4 -18x +12 = 36 add together, 12's cancel 10x = 40 x = 4 sub x's value to find y 7(4) - 3y = 1 28 - 3y = 1 27 -3y = 0 27 = 3y y = 9 final answer (4, 9)
Solve using elimination: -4y + 2x − 8 = 0 -3x + 6y = 12 Infinitely many solutions No solution (3, -1) (-1, 4)
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