Soru

Zorluk: OrtaSystems of Linear Equations

Consider the system of linear equations below.

3x4y=72x+3y=16\begin{aligned} 3x - 4y &= 7 \\ 2x + 3y &= 16 \end{aligned}

If (x,y)(x, y) is the solution to the system of equations above, what is the value of x+yx + y?

  1. 7Cevap
  2. B
    6
  3. C
    14
  4. D
    -9

Cevap

7
The correct answer is 7. By multiplying the first equation by 3 and the second by 4, we get 9x12y=219x - 12y = 21 and 8x+12y=648x + 12y = 64. Adding these equations eliminates the yy terms and gives 17x=8517x = 85, which simplifies to x=5x = 5. Substituting x=5x = 5 into the second equation yields 2(5)+3y=16    10+3y=16    3y=6    y=22(5) + 3y = 16 \implies 10 + 3y = 16 \implies 3y = 6 \implies y = 2. The sum of the coordinates is x+y=5+2=7x + y = 5 + 2 = 7.

Adım Adım Çözüm

1
Multiply the equations by constants to align the coefficients of one variable for elimination.
Multiplying the first equation by 3 gives 9x12y=219x - 12y = 21. Multiplying the second equation by 4 gives 8x+12y=648x + 12y = 64.
This sets up the system so that the yy terms have opposite coefficients (12-12 and +12+12), allowing them to be eliminated by addition.
2
Add the two modified equations to eliminate the yy variable and solve for xx.
(9x12y)+(8x+12y)=21+64    17x=85    x=5(9x - 12y) + (8x + 12y) = 21 + 64 \implies 17x = 85 \implies x = 5.
Adding the equations eliminates yy and leaves a single linear equation in terms of xx.
3
Substitute the value of xx back into one of the original equations to solve for yy.
Substituting x=5x = 5 into the second equation 2x+3y=162x + 3y = 16 gives 2(5)+3y=16    10+3y=16    3y=6    y=22(5) + 3y = 16 \implies 10 + 3y = 16 \implies 3y = 6 \implies y = 2.
Plugging the known variable value back in allows us to solve for the remaining unknown variable.
4
Calculate the value of the requested expression x+yx + y.
x+y=5+2=7x + y = 5 + 2 = 7.
The question specifically asks for the sum of xx and yy.

Anahtar Kavram

Solving systems of linear equations using the elimination method and evaluating linear combinations of the solutions.

Alternatif Yöntem

Alternatively, you can solve the first equation for xx in terms of yy: 3x=7+4y    x=7+4y33x = 7 + 4y \implies x = \frac{7+4y}{3}. Substitute this expression into the second equation: 2(7+4y3)+3y=162\left(\frac{7+4y}{3}\right) + 3y = 16. Multiply the entire equation by 3 to clear the fraction: 2(7+4y)+9y=48    14+8y+9y=48    17y=34    y=22(7+4y) + 9y = 48 \implies 14 + 8y + 9y = 48 \implies 17y = 34 \implies y = 2. Substitute y=2y = 2 back to find xx: x=7+4(2)3=5x = \frac{7+4(2)}{3} = 5. Thus, x+y=5+2=7x + y = 5 + 2 = 7.
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